Study Companion: S31 — Potential Energy and Stability

S30 found equilibrium positions from the virtual work of the active forces. Two of those forces are special. The work a spring or a weight does is not lost: it is stored, and it comes back when the motion reverses. Writing that work as the change of a stored quantity, the potential energy, turns the principle of virtual work into a statement about one function of position. Its slope is zero at equilibrium, and its curvature says something the force equations never did: whether the equilibrium will survive a small disturbance.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, Chapter 7, article 7/4 ("Potential Energy and Stability"), with Sample Problems 7/4–7/6; Marghitu, Dupac and Madsen, Statics with MATLAB, sections 6.2–6.4 ("Elastic Potential Energy", "Gravitational Potential Energy", "Stability of Equilibrium") and Examples 6.1–6.5; Harper, Solving Statics Problems in MATLAB, "Potential Energy" (its Sample Problem 7/5, the same in Meriam's 9th edition) and "Potential Energy and Stability" (its Problem 7/74, which is Meriam 9th edition Problem 7/63). Units SI throughout. Heights h are measured up from a stated datum, spring deformations x from the spring's free length, and angles in the sense marked on each figure.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

The owner's switchgear crank of S04 and S10: a crank of radius 60\,\mathrm{mm} on a shaft O, and a spring of stiffness 20\,\mathrm{kN/m} anchored at S, 240\,\mathrm{mm} from the shaft, pulling the crank pin P. A latch holds the crank at \varphi = 60^\circ, where the spring pulls 500\,\mathrm N (stretched 25\,\mathrm{mm}) and turns the crank with 22.7\,\mathrm{N\cdot m}. Turn the crank to \varphi = 0 and the pin lies on the line SO beyond the shaft: dead centre.

At dead centre the spring is at its longest, 300\,\mathrm{mm}, stretched about 50\,\mathrm{mm} and pulling about 1000\,\mathrm N, twice as hard as at the latch. Yet its line passes through the shaft, so its moment is zero: the crank is in equilibrium there. Release it exactly at dead centre: does it stay? The plausible answer is "yes, and firmly: the spring pulls hardest there and nothing turns it". The answer is in Nugget 06.

Nugget 01: A Spring Stores the Work Done on It

Claim. The work done in deforming a linear spring by x from its free length is stored in it as elastic potential energy V_e = \frac12kx^2, the same in tension or compression, and given back as the spring relaxes.

The picture. Meriam's Fig. 7/11: the force F = kx needed to hold the spring at deformation x.

Deforming the spring by dx at force F costs F\,dx, a thin strip under the line. Adding the strips from 0 to x fills the triangle, base x, height kx: area \frac12kx^2. From x_1 to x_2 the strips fill the trapezoid between them.

Check with numbers. The switchgear spring, k = 20\,\mathrm{kN/m}: stretched 25\,\mathrm{mm} it holds \frac12(20\,000)(0.025)^2 = 6.25\,\mathrm J; stretched 50\,\mathrm{mm}, 25.0\,\mathrm J. Going from 25 to 50\,\mathrm{mm} adds the trapezoid, \frac12(20\,000)(0.050^2 - 0.025^2) = 18.75\,\mathrm J: four times the energy for twice the stretch.

The idea. Meriam (7/4) and Marghitu (6.2): for a linear elastic spring, F = kx, and the work done on it during dx is dU = F\,dx, so

The same holds for a spring in tension with x its elongation. During relaxation the change, final minus initial, is negative. The force on the body at the spring's moving end is the negative of the force on the spring, so the work done on the body by the spring is -\Delta V_e. A torsional spring of stiffness k_T (torque per radian) twisted by \theta resists with M = k_T\theta and stores

in joules. Marghitu (6.2) adds how springs combine. In parallel they share one deformation and their forces add, so k = k_1 + k_2 + \dots (6.2); in series they carry one force and their deformations add, so 1/k = 1/k_1 + 1/k_2 + \dots (6.3).

The equation (material model, linear spring, and its stored energy):

Assumptions. Linear elastic spring; x measured from the free length; massless spring.

Worked number. A torsion spring of 1.8\,\mathrm{N\cdot m/rad} wound a quarter turn stores \frac12(1.8)(\pi/2)^2 = 2.22\,\mathrm J. Two 20\,\mathrm{kN/m} springs side by side are 40\,\mathrm{kN/m} (25\,\mathrm{mm} needs 500 + 500\,\mathrm N); end to end, 10\,\mathrm{kN/m} (500\,\mathrm N stretches each 25\,\mathrm{mm}, 50\,\mathrm{mm} in all).

Real example. A spring-operated circuit breaker stores its closing energy exactly so: the charging motor stretches the spring, and the 18.75\,\mathrm J between two stretches is what the spring can give back to drive the contacts.

Plausible wrong answer. "Energy is force times stretch: 500\,\mathrm N \times 25\,\mathrm{mm} = 12.5\,\mathrm J." The force grew from zero as the spring stretched; the average is half the final force, so 6.25\,\mathrm J.

Check. A spring stretched x is stretched to 2x. By what factor does its stored energy grow? (Four.)

Narration spine. A spring stretches as a point climbs the line F = kx; thin strips stack under it into a blue triangle reading 6.25\,\mathrm J. The stretch doubles and an orange trapezoid adds 18.75\,\mathrm J. A torsion spring winds beside it with \frac12k_T\theta^2.

Bridge. A raised weight stores energy too, and its account is even simpler.

Nugget 02: Lifting Stores mgh, Whatever the Path

Claim. A body whose mass centre is a height h above a chosen datum has gravitational potential energy V_g = mgh. Only changes in it matter, so the datum is arbitrary, and the change depends only on the change of height, not on the path.

The picture. Meriam's Figs. 7/12 and 7/13.

Left: above the datum V_g = +mgh, below it -mgh. Right: three paths from one level to another \Delta h higher. The weight is vertical, so only the vertical part of each step does work: every path costs mg\,\Delta h.

Check with numbers. Harper's 25\,\mathrm{kg} trap door has its centre 0.3\,\mathrm m from the hinge. Raised from flat to vertical: V_g = 25(9.81)(0.3) = 73.6\,\mathrm J; at 30^\circ, 73.575\sin30^\circ = 36.8\,\mathrm J. Put the datum 1\,\mathrm m lower: the values become 245.25\,\mathrm J and 318.8\,\mathrm J, and their difference is still 73.6\,\mathrm J.

The idea. Meriam (7/4) and Marghitu (6.3): S30 treated the weight as an active force: in an upward virtual displacement \delta h it does \delta U = -mg\,\delta h. Alternatively define the gravitational potential energy as the work done against the weight, by a force equal and opposite to it, in bringing the body from the datum to its position: the negative of the weight's work. Then

with \delta h the upward virtual displacement of the mass centre (\delta V_g negative if it moves down). Marghitu shows the path independence directly: along any path the work of the weight is \int G\cos\theta\,ds = \int mg\,dy, which depends only on the end heights. The unit is the joule.

The equation (definition, gravitational potential energy in a uniform field):

Worked number. A 60\,\mathrm{kg} crate lifted 1.2\,\mathrm m gains 60(9.81)(1.2) = 706.3\,\mathrm J, whether it goes straight up, up a ramp or round a spiral stair.

Real example. Garage doors and trap doors are balanced with springs so that the gain in V_g as the door rises is matched by the loss in V_e of the springs, and the door stays where it is put. Nugget 07 asks how well that works.

Plausible wrong answer. "A body below the datum has no potential energy." It has negative V_g: the datum is only where the count starts.

Check. Does moving the datum change where a system is in equilibrium? (No: it adds a constant to V, which has no slope.)

Narration spine. A block hovers above a datum line labelled +mgh; another below reads -mgh. The datum slides down and both labels change by the same amount. Three coloured paths climb from G_1 to G_2; all three counters stop at mg\,\Delta h.

Bridge. Springs and weights are not the only forces with a potential. Which forces have one, and how can that be told?

Nugget 03: A Force Has a Potential Only If Its Work Is Path-Independent

Claim. A force has a potential energy V when its work between two positions is the same along every path; then dV = -\mathbf F\cdot d\mathbf r, and the force is called conservative. A force field whose curl is not zero does work round a closed loop and has no potential.

The picture. Left: weight or a spring force, two paths, the same work. Right: a force that circulates.

If two paths from A to B gave different works, going out on one and back on the other would do net work round a loop, and no single number V(A) - V(B) could describe it. The circulating force does exactly that.

Check with numbers. The force \mathbf F = -y\mathbf i + x\mathbf j\,\mathrm N (x, y in m), taken counter-clockwise round the unit square: along the bottom (y = 0) F_x = 0; up the right side (x = 1) F_y = 1, work 1\,\mathrm J; along the top (y = 1, moving left) F_x = -1, work 1\,\mathrm J; down the left (x = 0) F_y = 0. Total 2\,\mathrm J \ne 0. Its curl, \partial F_y/\partial x - \partial F_x/\partial y = 1 - (-1) = 2, is not zero.

The idea. Marghitu (6.2): a function V of position is the potential energy associated with a force \mathbf F if dV = -\mathbf F\cdot d\mathbf r = -dU for every d\mathbf r; such a force is conservative, and a spring's force is one. Marghitu tests a force by its curl, \nabla\times\mathbf F = \mathbf 0, and finds V = -\int\mathbf F\cdot d\mathbf r. Example 6.3: a particle on the helix x = a\cos\theta, y = a\sin\theta, z = b\theta, pulled towards the origin by \mathbf F = -k\mathbf r and by its weight. The total force has zero curl, and V = \frac12k(x^2 + y^2 + z^2) + mgz = \frac12k(a^2 + b^2\theta^2) + mgb\theta, so dV/d\theta = mgb + kb^2\theta = 0 at z = b\theta = -mg/k, and d^2V/d\theta^2 = kb^2 > 0. Example 6.4: a particle on a sphere of radius R (centre O) is pulled by k times its distance towards the top point A(0, 0, R); again the curl is zero. Kinetic friction, which always opposes the motion, has no potential (S30).

The equation (definition: conservative force and its potential):

Worked number. Example 6.3 with a = 3\,\mathrm m, b = 0.5\,\mathrm m, k = 10\,\mathrm{N/m}, m = 1\,\mathrm{kg}: z = -9.81/10 = -0.981\,\mathrm m, \theta = -1.962\,\mathrm{rad} (-112.4^\circ), at (-1.144, -2.773, -0.981)\,\mathrm m (Marghitu, using g = 9.8: -0.980\,\mathrm m, -1.960\,\mathrm{rad}, (-1.138, -2.776)). Example 6.4 on the sphere: x^2 + y^2 = R^2 - z^2 turns V into kR^2 + (mg - kR)z, a straight line in z, so the poles are the equilibria (Marghitu's result). Ours: with R = 0.5\,\mathrm m, m = 1\,\mathrm{kg}, k = 10\,\mathrm{N/m}, kR = 5\,\mathrm N < mg, so V is lowest at the bottom pole (stable) and highest at the top (unstable).

Real example. Every potential-energy calculation in this lesson rests on this test: weights and springs pass it; friction, and a hand pushing a lever, do not, and stay as active forces.

Plausible wrong answer. "Any force that depends only on position has a potential." The circulating force (-y, x) depends only on position and still does 2\,\mathrm J round a loop.

Check. Is \mathbf F = 2xy\,\mathbf i + x^2\,\mathbf j conservative? (Yes: \partial(x^2)/\partial x - \partial(2xy)/\partial y = 2x - 2x = 0; V = -x^2y.)

Narration spine. Two curves join A to B and two work counters reach the same number. On the right, arrows swirl round a square; a dot goes once round and its counter ends at 2\,\mathrm J, not zero. The word "conservative" settles on the left.

Bridge. With springs and weights described by V, the principle of virtual work can be rewritten.

Nugget 04: Springs and Weights Move Into the Energy: Virtual Work Equals the Change in V

Claim. If the springs are drawn inside the system and the weights are accounted for by V_g, the principle of virtual work becomes \delta U' = \delta V: the virtual work of the remaining active forces equals the virtual change in total potential energy.

The picture. Meriam's Fig. 7/14: a particle on a smooth guide, with a spring, its weight and two applied forces.

Left: the particle alone. All four forces, F_1, F_2, mg and the spring force kx, are active, and \delta U = 0 (the guide's normal force does no work). Right: the spring drawn inside the boundary. Its forces on the particle and on the ground are now internal and not drawn; its work, and the weight's, are counted in \delta V. Only F_1 and F_2 remain, doing \delta U'.

Check with numbers. Meriam's Sample Problem 7/5: two uniform links of mass m and length 2b hang from O and from a collar C; a light rod from A passes through a pivoted collar at B (each b from the top along its link) and compresses a spring k that is free at \theta = 0. A moves away from B by x = 2b\sin(\theta/2), so V_e = \frac12k(2b\sin\frac{\theta}{2})^2 = 2kb^2\sin^2\frac{\theta}{2}. With the datum through O, each centre is b\cos(\theta/2) below: V_g = -2mgb\cos\frac{\theta}{2}. C is 4b\sin(\theta/2) from O, so \delta U' = P\,\delta(4b\sin\frac{\theta}{2}) = 2Pb\cos\frac{\theta}{2}\,\delta\theta. Then 2Pb\cos\frac{\theta}{2}\,\delta\theta = 2kb^2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\,\delta\theta + mgb\sin\frac{\theta}{2}\,\delta\theta, and

The idea. Meriam (7/4) and Marghitu (6.3): the work done on a body by a spring is -\delta V_e and by its weight -\delta V_g. Split the total virtual work into \delta U', done by all active forces other than springs and weights, and -(\delta V_e + \delta V_g); then Eq. 7/3 becomes

with V = V_e + V_g (Marghitu writes \delta U_a = \delta V). The virtual work done by all external active forces, other than the gravitational and spring forces accounted for in the potential energy terms, on a mechanical system in equilibrium equals the corresponding change in the total elastic and gravitational potential energy, for any and all virtual displacements consistent with the constraints. The active-force diagram must make clear what is inside the boundary.

The equation (physical law, the principle of virtual work restated with potential energy):

Assumptions. Ideal system as in S30; linear springs; uniform gravity.

Worked number. Sample Problem 7/5 cannot be solved for \theta explicitly (Meriam); Harper does it numerically. With his data in SI, P = 100\,\mathrm{lb} = 444.8\,\mathrm N, mg = 50\,\mathrm{lb} = 222.4\,\mathrm N, k = 20\,\mathrm{lb/in} = 3502.5\,\mathrm{N/m}, b = 5\,\mathrm{in} = 0.127\,\mathrm m: kb = 444.8\,\mathrm N, and dividing by mg leaves one parameter, P/mg = \beta\sin\frac{\theta}{2} + \frac12\tan\frac{\theta}{2} with \beta = kb/mg = 2. Solving 2 = 2\sin\frac{\theta}{2} + \frac12\tan\frac{\theta}{2} gives \theta = 1.6415\,\mathrm{rad} = 94.05^\circ (Harper's graph: about 95^\circ). Harper's second case, P/mg = 2.2 with \beta = 1.74, he reads as about 112^\circ; the root is 112.7^\circ. Marghitu's Example 6.1 runs the other way: two 1\,\mathrm{kg} links of 5\,\mathrm m, a 500\,\mathrm N force and a spring stretched x = 2l(1 - \cos\theta); for equilibrium at \theta = 30^\circ, k = (2F\sin\theta + mg\cos\theta)/(4l(1 - \cos\theta)\sin\theta) = 379.5\,\mathrm{N/m}.

Real example. Spring-balanced mechanisms (lift platforms, folding beds, the switchgear's charging gear) are designed with \delta U' = \delta V: one equation in the position, with every pin and reaction absent.

Plausible wrong answer. "Put the spring force in \delta U' and also its energy in \delta V." That counts the spring twice. Either the spring is outside the system (its force is active, Eq. 7/3) or inside (its energy is in V, Eq. 7/6).

Check. In Sample Problem 7/5, why does P grow with both k and m? (Opening the links compresses the spring and raises both centres, so both energies grow with \theta.)

Narration spine. The particle on its guide shows four arrows; a dashed boundary swallows the spring, the kx and mg arrows fade into a "\delta V" label, and F_1, F_2 remain under "\delta U'". Cut to SP 7/5: the links open, the spring shortens, a P(\theta) curve rises and a line at P/mg = 2 cuts it at 94^\circ.

Bridge. When no force outside the potential acts at all, \delta U' = 0, and equilibrium becomes a statement about V alone.

Nugget 05: With No Other Forces, Equilibrium Is Where V Is Flat

Claim. If only springs and weights do work, \delta V = 0: a system is in equilibrium exactly where its total potential energy is stationary, dV/dx = 0 for its one coordinate x.

The picture. Meriam's Sample Problem 7/4: a 10\,\mathrm{kg} cylinder hangs on a spring of 2\,\mathrm{kN/m}; x is measured down from the spring's free length, with V = 0 there.

The spring's energy \frac12kx^2 is a parabola rising from zero; the weight's, -mgx, a line falling steadily. Their sum dips to a minimum and climbs again. At the bottom of the dip a small move either way changes V by a second-order amount only: the slope is zero.

Check with numbers. V = \frac12(2000)x^2 - 10(9.81)x. dV/dx = 2000x - 98.1 = 0 at x = 0.04905\,\mathrm m = 49.05\,\mathrm{mm} (Meriam prints 0.0490\,\mathrm m, 49.0\,\mathrm{mm}, a rounding of the same value). There V = 2.406 - 4.812 = -2.41\,\mathrm J. At x = 0.10\,\mathrm m, V = 10 - 9.81 = 0.19\,\mathrm J. And kx = 98.1\,\mathrm N = mg: the spring holds the weight, as the force balance says.

The idea. Meriam (7/4), with \delta U' = 0 Eq. 7/6 becomes

The equilibrium configuration is one where V has a stationary value. For one degree of freedom x, with V and its derivatives continuous, \delta V = (dV/dx)\,\delta x = 0 for any \delta x, so

With several degrees of freedom each partial derivative must vanish. Marghitu (6.4) writes the same with a generalised coordinate q, Eqs. 6.5–6.7, and calls \delta V = 0 the principle of virtual work for conservative forces. Its Example 6.5: a particle on a vertical circular wire of radius R, V = mgR\sin\theta; dV/d\theta = mgR\cos\theta = 0 at the top (\pi/2) and the bottom (3\pi/2).

The equation (mathematical identity for one degree of freedom, from the physical law 7/7):

Assumptions. Only conservative forces do work (\delta U' = 0); V differentiable.

Worked number. Marghitu's Example 6.5 with R = 0.2\,\mathrm m, m = 0.5\,\mathrm{kg}: d^2V/d\theta^2 = -mgR\sin\theta is -0.981\,\mathrm{J/rad^2} at the top and +0.981 at the bottom (sorted in Nugget 06).

Real example. Meriam's hint to SP 7/4: moving the datum of V_e or V_g shifts the curves up or down but never moves the minimum, so any convenient datum will do.

Plausible wrong answer. "Equilibrium is where V is zero." V is zero wherever the data put it; the equilibrium is where V is flat, here at V = -2.41\,\mathrm J.

Check. At what x does V return to zero in SP 7/4? (At x = 2mg/k = 98.1\,\mathrm{mm}, twice the equilibrium stretch.)

Narration spine. The cylinder bobs on its spring beside a plot. A blue parabola and an orange line are drawn, then added into a green curve; a tangent slides along it and turns horizontal at 49\,\mathrm{mm}, where the cylinder comes to rest.

Bridge. A stationary V can be a valley, a peak or a plateau. They are all equilibria, but they behave very differently when nudged.

Nugget 06: The Second Derivative Decides: Stable, Unstable or Neutral

Claim. An equilibrium is stable where V is a minimum (d^2V/dx^2 > 0), unstable where it is a maximum (d^2V/dx^2 < 0) and neutral where V is constant. If the second derivative is zero, the first non-zero higher derivative decides.

The picture. Left: Meriam's Fig. 7/15. Right: the switchgear spring's potential energy against the crank angle.

A ball in a dish sits at a minimum of V: a push raises its energy and it rolls back. On a dome it sits at a maximum: a push lowers its energy and it rolls away. On a flat plane any position is as good as any other. The crank's curve is a hill with dead centre on its top.

Check with numbers. The puzzle. The spring's length is PS = \sqrt{r^2 + s^2 + 2rs\cos\varphi} (r = 0.06, s = 0.24\,\mathrm m), 300\,\mathrm{mm} at \varphi = 0, and its free length is 249.95\,\mathrm{mm} (from 500\,\mathrm N at 275\,\mathrm{mm}). V = \frac12k(PS - 0.24995)^2: 25.05\,\mathrm J at dead centre, 6.25\,\mathrm J at the latch. dV/d\varphi = k(PS - L_0)\,d(PS)/d\varphi, and d(PS)/d\varphi = -rs\sin\varphi/PS is zero at \varphi = 0: equilibrium, as the zero moment said. The second derivative there is k(PS - L_0)(-rs/PS) = 20\,000(0.05005)(-0.048) = -48.05\,\mathrm{N\cdot m/rad}. Negative: dead centre is unstable. One degree off, the spring already drives the crank away with 0.84\,\mathrm{N\cdot m}, and it snaps to the latch, releasing 25.05 - 6.25 = 18.80\,\mathrm J. Pulling hardest is not holding firmest.

The idea. Meriam (7/4) and Marghitu (6.4): Eq. 7/8 holds when V is a minimum (stable equilibrium), a maximum (unstable) or constant (neutral). A small displacement from a stable position increases V and the system tends back to lower energy; from an unstable one it decreases V and the system moves farther away; in a neutral one nothing changes. For a function with continuous derivatives the second derivative is positive at a minimum and negative at a maximum, so

If d^2V/dx^2 = 0, look at higher derivatives: if the lowest non-zero one is of even order, the equilibrium is stable or unstable as its sign is positive or negative; if it is of odd order, the equilibrium is unstable, and V has an inflection with zero slope there (Meriam). Several degrees of freedom need more (a Taylor expansion in two variables for two).

The equation (mathematical identity: the second-derivative test, applied to V):

Worked number. Meriam's Sample Problem 7/6: a uniform bar of mass m and length b with its ends in a vertical and a horizontal guide; a spring along the horizontal guide, free when the bar is vertical (x = b\sin\theta = 0). With the horizontal guide as datum, V = \frac12kb^2\sin^2\theta + \frac12mgb\cos\theta, so dV/d\theta = (kb^2\cos\theta - \frac12mgb)\sin\theta = 0 at \theta = 0 and at \cos\theta = mg/2kb. Then d^2V/d\theta^2 = kb^2(2\cos^2\theta - 1) - \frac12mgb\cos\theta: at \theta = 0 it is kb^2(1 - mg/2kb), positive (stable) if k > mg/2b; at the second position it is kb^2[(mg/2kb)^2 - 1] < 0, never stable. With m = 2\,\mathrm{kg}, b = 0.6\,\mathrm m: mg/2b = 16.35\,\mathrm{N/m}; for k = 50\,\mathrm{N/m} the vertical position is stable (+12.11\,\mathrm{J/rad^2}) though the spring carries no force there, and \theta = 70.91^\circ is an unstable equilibrium (-16.08\,\mathrm{J/rad^2}). Marghitu's Example 6.2 (a bar with its top on a vertical spring, l = 5\,\mathrm m, k = 100\,\mathrm{N/m}, m = 10\,\mathrm{kg}, g = 9.8): \theta = 0 unstable (-mgl/2 = -245), \theta = \pm25.6^\circ stable (mg(l - mg/4k) = 466).

Real example. Toggle switches, over-centre clamps and the switchgear charging spring are built round an unstable dead centre: the energy hill is what makes them snap decisively to one side instead of creeping. An inverted pendulum stays up only if springs make V a minimum there (Self-check 9).

Plausible wrong answer. "Zero force (or zero moment) means it will stay." That only makes it an equilibrium; staying needs V to be a minimum.

Check. V = x^4 at x = 0: stable or not? (d^2V/dx^2 = 0, d^3V/dx^3 = 0, d^4V/dx^4 = 24 > 0: even order, positive, stable.)

Narration spine. Three balls on a dish, a dome and a plane are nudged: one returns, one rolls off, one drifts. The crank appears with its energy hill; the crank sits on the summit at dead centre, is nudged one degree, and snaps to the latch as a dot runs down the hill.

Bridge. In the bar of SP 7/6 the verdict at \theta = 0 flipped as k passed mg/2b. Harper's trap door shows that flip as a whole curve.

Nugget 07: A Parameter Can Flip the Verdict: the Trap Door

Claim. Changing a stiffness or a mass can turn a stable equilibrium unstable. The switch happens where the second derivative passes through zero, at a critical value of the parameter, and there the higher derivatives decide.

The picture. Harper's trap door (Meriam's Problem 7/63): a uniform 25\,\mathrm{kg} door, 0.6\,\mathrm m from its hinge O–O to its top edge, held by two springs from its top corners to floor points 0.6\,\mathrm m beyond the hinge. The springs are free at \theta = 90^\circ (door vertical). V against \theta for three stiffnesses:

All three curves cross at 90^\circ, where the springs are free and V = V_g = 73.6\,\mathrm J; there each is flat. For k = 100\,\mathrm{N/m} the crossing is a peak, for 400\,\mathrm{N/m} a valley, and for 204.4\,\mathrm{N/m} neither.

Check with numbers. By the law of cosines the spring length is AB = \sqrt{0.72(1 + \cos\theta)}\,\mathrm m, free length 0.6\sqrt2 = 0.8485\,\mathrm m, so for the two springs V = 25(9.81)(0.3)\sin\theta + k(AB - 0.8485)^2 (Harper). k = 100: equilibria at 90^\circ (unstable) and 152.0^\circ (stable). k = 400: 48.8^\circ (unstable) and 90^\circ (stable). At 90^\circ, d^2V/d\theta^2 = -73.575 + 0.36k, which is zero at k = 204.375\,\mathrm{N/m}, Harper's critical stiffness.

The idea. Harper (7.3): a change in a system's behaviour as a parameter varies is a bifurcation, and the value where it happens a critical point, found by setting the second derivative to zero. At 90^\circ the door is in equilibrium for every k: with no springs it is an inverted pendulum, unstable; stiff enough springs hold it. Two points need care. (1) At the critical k the second derivative is zero and the third is about 110\,\mathrm{J/rad^3}: Harper calls this "half-stable" (energy rises to one side, falls to the other); by Meriam's rule, an odd lowest order, it is unstable, which is the verdict used here. (2) One-sided springs. Past 90^\circ the springs are shorter than their free length, and Harper's V lets them push. The "stable" 152.0^\circ at k = 100 needs each spring compressed by 558\,\mathrm{mm} of its 849\,\mathrm{mm} free length. Extension springs go slack instead; then past 90^\circ only the weight acts, V falls, and the door falls open. So with real extension springs the k = 400 door is held at 90^\circ only against pushes towards smaller \theta. Meriam's version of the problem restricts \theta to 0–90^\circ, where the springs are always stretched.

The equation (geometric relationship plus potential energy, for this door):

Assumptions. Springs linear and acting in tension and compression as Harper assumes (see point 2); hinge frictionless; door uniform.

Worked number. Meriam's 9th-edition version, k = 800\,\mathrm{N/m} over 0 to 90^\circ: dV/d\theta = 0 at \theta = 24.8^\circ, where d^2V/d\theta^2 = -155.9\,\mathrm{J/rad^2}: unstable, as Meriam's answer states. Released just above 24.8^\circ the door rises to vertical; just below, it falls flat.

Real example. Every spring-balanced door, hatch and boom has such a critical stiffness. Below it the balanced position is a peak and the door slams; well above it the door is held, but needs a push to move. Designers aim near the critical value, so that V is almost flat and the door stays wherever it is left (neutral, in the limit).

Plausible wrong answer. "An equilibrium that exists for every k has the same stability for every k." The trap door is in equilibrium at 90^\circ for all k, unstable below 204.4\,\mathrm{N/m} and stable above.

Check. Why do all three curves pass through the same point at 90^\circ? (The springs are at their free length, so V_e = 0 for every k, and V = V_g = 73.6\,\mathrm J.)

Narration spine. The door swings on its hinge with a stiffness slider beside it. At k = 100 the V curve has a peak at 90^\circ and the door topples; as k rises the peak flattens, and at 204.4 the curve has a level inflection; at 400 it is a valley and the door stands. The region past 90^\circ greys out with "springs compressed".

Synthesis

The switchgear crank is in equilibrium at dead centre, because the spring's line passes through the shaft, but it cannot stay there: the spring's energy, 25.05\,\mathrm J, is at the top of a hill, and the second derivative, -48\,\mathrm{N\cdot m/rad}, sends it off at the slightest touch. Equilibrium was a statement about forces; stability is a statement about the shape of the energy. For a conservative system with one degree of freedom, equilibrium is where dV/dx = 0; it is stable where V is a minimum, unstable where it is a maximum, and neutral where it is constant. Springs (\frac12kx^2) and weights (mgh) supply V; other active forces stay in \delta U' = \delta V.

• Central idea: the work of springs and weights is stored; equilibrium is a stationary point of V.
• Key relationship: \delta U' = \delta V; with \delta U' = 0, dV/dx = 0, and the sign of d^2V/dx^2 (or the first non-zero higher derivative) gives the type.
• Most important assumption: the forces in V are conservative and the springs act as modelled (an extension spring cannot push; a cable cannot push).
• Open question: with two or more degrees of freedom the test needs every second partial derivative together: how does a Taylor expansion in two variables decide stability?

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Marghitu's and Harper's programs; the Python is a translation, and both print the same numbers.

Python (numpy, scipy, sympy). S31_potential_energy.py writes V(\theta), solves dV/d\theta = 0 with sympy, and reads the sign of d^2V/d\theta^2 at each root, for Sample Problem 7/6 and Marghitu's Example 6.2. It solves Harper's Sample Problem 7/5, which has no explicit answer, with brentq, and checks Marghitu's Examples 6.1 and 6.3.

Part 1: stability from V(theta), with sympy
  SP 7/6: V = b**2*k*sin(theta)**2/2 + b*g*m*cos(theta)/2
  dV/dtheta = b*(2*b*k*cos(theta) - g*m)*sin(theta)/2
  m = 2 kg, b = 0.6 m, k = 50 N/m:
    theta =   0.00 deg: V'' =   +12.11 J/rad^2, stable
    theta =  70.91 deg: V'' =   -16.08 J/rad^2, unstable
  Marghitu Ex 6.2, l = 5 m, k = 100 N/m, m = 10 kg, g = 9.8:
    theta =   0.00 deg: V'' =  -245.00 J/rad^2, unstable
    theta =  25.58 deg: V'' =  +465.99 J/rad^2, stable
Part 2: Harper's SP 7/5, P / mg = beta sin(theta/2) + tan(theta/2) / 2
  P/mg = 2.0, beta = 2.0: theta = 1.6415 rad = 94.05 deg
  P/mg = 2.2, beta = 1.74: theta = 1.9673 rad = 112.72 deg
Part 3: Marghitu's examples and springs
  Ex 6.1: k for equilibrium at 30 deg = 379.5 N/m
  Ex 6.3 bead on the helix: z = -0.981 m, theta = -1.962 rad, at (-1.144, -2.773, -0.981) m
  torsion spring 1.8 N m/rad, quarter turn: 2.22 J
  crate 60 kg up 1.2 m: 706.3 J by any path

MATLAB (Octave-compatible). S31_potential_energy.m evaluates the second derivatives directly and solves Harper's equation with fzero.

SP 7/6: theta = 0, V'' = +12.11; theta = 70.91 deg, V'' = -16.08 J/rad^2
P/mg = 2.0, beta = 2.00: theta = 1.6415 rad = 94.05 deg
P/mg = 2.2, beta = 1.74: theta = 1.9673 rad = 112.72 deg
Ex 6.1: k = 379.5 N/m

Traps

• Energy as force times stretch. A spring's energy is \frac12kx^2, half of final force times stretch. (From Nugget 01.)
• Measuring x from the wrong place. x in \frac12kx^2 is the deformation from the free length, not the spring's length. (From Nugget 01.)
• Adding series springs like parallel ones. In series 1/k = \Sigma1/k_i; the result is softer than either. (From Nugget 01.)
• Using \delta h downward in \delta V_g = mg\,\delta h. \delta h is the upward move of the mass centre. (From Nugget 02.)
• Assuming any force has a potential. Check the curl; friction and circulating forces have none. (From Nugget 03.)
• Counting a spring twice. Its force is in \delta U' or its energy in \delta V, never both. (From Nugget 04.)
• Taking V = 0 for equilibrium. Equilibrium is where dV/dx = 0. (From Nugget 05.)
• Losing a root. \sin\theta\,(\dots) = 0 has the root \sin\theta = 0 as well (Meriam's SP 7/6 hint). (From Nugget 06.)
• Equilibrium taken for stability. Zero moment at dead centre does not hold the crank; check d^2V/dx^2. (From Nugget 06.)
• Letting a spring push past its free length. Check every equilibrium against the springs' one-sided behaviour. (From Nugget 07.)

Self-check

1. A spring of 5\,\mathrm{kN/m} is stretched from 10 to 40\,\mathrm{mm}. Find \Delta V_e, and the work the spring does on the body attached to it.
2. Springs of 3 and 6\,\mathrm{kN/m} are joined (a) side by side, (b) end to end. Find the combined stiffness.
3. A torsion spring of 5\,\mathrm{N\cdot m/rad} is wound 60^\circ. What energy does it store?
4. A 60\,\mathrm{kg} crate is lifted 1.2\,\mathrm m from a shelf 0.8\,\mathrm m above the floor. Find \Delta V_g with the datum at the shelf and with it at the floor.
5. Meriam's Problem 7/29: V = 6x^4 - 3x^2 + 5 (in J, x in m). Find the equilibrium positions and their stability.
6. Meriam's Sample Problem 7/4 with a 15\,\mathrm{kg} cylinder and a 2.5\,\mathrm{kN/m} spring: find the equilibrium x and the minimum V.
7. Which of \mathbf F = 2xy\,\mathbf i + x^2\,\mathbf j and \mathbf F = y\,\mathbf i - x\,\mathbf j is conservative? Give V for it.
8. Meriam's Problem 7/34: a uniform bar of mass m and length l, hinged at its lower end, carries a torsion spring M = k_T\theta (\theta from the vertical). Find the largest l for which \theta = 0 is stable, and evaluate it for m = 2\,\mathrm{kg}, k_T = 15\,\mathrm{N\cdot m/rad}.
9. Meriam's Problem 7/41: an inverted pendulum, a mass m at height h above a pivot, is steadied by two springs k at distances b either side, equally precompressed. Find the largest h for stability; evaluate for k = 5\,\mathrm{kN/m}, b = 0.2\,\mathrm m, m = 50\,\mathrm{kg}.
10. Meriam's Sample Problem 7/6 with m = 4\,\mathrm{kg}, b = 0.5\,\mathrm m, k = 100\,\mathrm{N/m}: find both equilibrium positions and their stability.
11. Meriam's Problem 7/30: two uniform bars OA and AB (each m = 1.25\,\mathrm{kg}, b = 0.75\,\mathrm m) are joined at A by a torsion spring k_T = 1.8\,\mathrm{N\cdot m/rad}, free when the bars are vertical and overlapping; O is pinned and B rolls on the floor, each bar at \theta to the floor. Find the equilibria in 0 \le \theta \le 90^\circ and their stability.
12. Meriam's Problem 7/63 (Harper's trap door with k = 800\,\mathrm{N/m}): find the equilibrium in 0 < \theta < 90^\circ and its stability.

Fully worked solutions

1. \Delta V_e = \frac12(5000)(0.040^2 - 0.010^2) = 2500 \times 0.0015 = 3.75\,\mathrm J. The spring does -3.75\,\mathrm J on the body.

2. (a) Same deformation, forces add: 3 + 6 = 9\,\mathrm{kN/m}. (b) Same force, deformations add: 1/k = 1/3 + 1/6 = 1/2, k = 2\,\mathrm{kN/m}.

3. \theta = \pi/3 = 1.047\,\mathrm{rad}; V_e = \frac12(5)(1.047)^2 = 2.74\,\mathrm J.

4. Shelf datum: 0 \to 60(9.81)(1.2) = 706.3\,\mathrm J. Floor datum: 60(9.81)(0.8) = 470.9\,\mathrm J \to 60(9.81)(2.0) = 1177.2\,\mathrm J, a change of 706.3\,\mathrm J. Same.

5. dV/dx = 24x^3 - 6x = 6x(4x^2 - 1) = 0: x = 0, x = \pm\frac12\,\mathrm m. d^2V/dx^2 = 72x^2 - 6: -6 at x = 0, unstable; 72(\frac14) - 6 = 12 at x = \pm\frac12, stable (Meriam's answer), with V = 4.625\,\mathrm J there.

6. x = mg/k = 147.15/2500 = 0.05886\,\mathrm m = 58.86\,\mathrm{mm}; V_{\min} = \frac12kx^2 - mgx = -(mg)^2/2k = -147.15^2/5000 = -4.33\,\mathrm J; d^2V/dx^2 = k > 0, stable.

7. First: \partial(x^2)/\partial x - \partial(2xy)/\partial y = 2x - 2x = 0, conservative, V = -x^2y (check: -\partial V/\partial x = 2xy, -\partial V/\partial y = x^2). Second: \partial(-x)/\partial x - \partial(y)/\partial y = -2 \ne 0, not conservative.

8. V = \frac12k_T\theta^2 + mg\frac l2\cos\theta; dV/d\theta = k_T\theta - \frac12mgl\sin\theta = 0 at \theta = 0; d^2V/d\theta^2 = k_T - \frac12mgl\cos\theta = k_T - \frac12mgl at 0. Stable if l < 2k_T/mg = 30/19.62 = 1.53\,\mathrm m.

9. Tilt \theta: one spring is compressed b\sin\theta more, the other b\sin\theta less; at \theta = 0 the precompression terms cancel in dV/d\theta, and d^2V/d\theta^2 = 2kb^2 - mgh. Stable if h < 2kb^2/mg = 2(5000)(0.04)/490.5 = 0.816\,\mathrm m.

10. mg/2b = 39.24/1 = 39.24\,\mathrm{N/m} < k: \theta = 0 stable, d^2V/d\theta^2 = kb^2(1 - mg/2kb) = 25(1 - 0.3924) = 15.19\,\mathrm{J/rad^2}. Second position \cos\theta = mg/2kb = 0.3924, \theta = 66.90^\circ: kb^2(0.3924^2 - 1) = -21.15\,\mathrm{J/rad^2}, unstable.

11. The spring's twist is the angle between the bars, \pi - 2\theta; each centre is \frac b2\sin\theta high. V = \frac12k_T(\pi - 2\theta)^2 + mgb\sin\theta; dV/d\theta = -2k_T(\pi - 2\theta) + mgb\cos\theta, with mgb = 9.197\,\mathrm{N\cdot m}. It vanishes at \theta = 90^\circ (0 + 0) and, solving numerically, at \theta = 22.3^\circ. d^2V/d\theta^2 = 4k_T - mgb\sin\theta: at 22.3^\circ, 7.2 - 3.49 = 3.71 > 0, stable; at 90^\circ, 7.2 - 9.197 = -2.00 < 0, unstable (Meriam's answer).

12. V = 73.575\sin\theta + 800\big(\sqrt{0.72(1 + \cos\theta)} - 0.8485\big)^2; dV/d\theta = 0 at \theta = 24.8^\circ, where d^2V/d\theta^2 = -155.9\,\mathrm{J/rad^2}: unstable (Meriam: 24.8^\circ, unstable). The springs are stretched throughout 0–90^\circ, so the model holds.