When a chain collides with a wall or floor it can get pulled in. Various mechanical implementations show the idea. It shows that the class of classic chain lifting and falling problems are actually generally ill-posed.

Linear 2-D stiffness matrices have orthogonal principal axes. If they arise from 3 or more symmetrically distributed springs, the fact that each spring represents an axis of symmetry means that the principal stiffnesses are equal. Therefore two equal springs at right angles also creates isotropic support.

The key insight is that when a half-space is loaded over a convex 'patch', deformation is concentrated under the patch in a block of material as deep as the patch width (minor diameter). The proper Hertz formulae appear when vertically uniform compression at the center of the blocks (in both bodies) suffices to let the bodies contact at the edges of the patch.

Contact, even in the context of linear elasticity, is famously non-linear. Hertz contact for example gives a displacement that is a non-linear power law function of applied load. Here is a contact problem that is linear in a way: the deformation of a structure where two bodies have, when unloaded, fully conformal contact but where the contact region may diminish when load is applied. Example: put a flat bottom beam straight beam on a rigid support surface. With no load the beam can contact the rigid support along its full length and width. Now press down on the beam with some load distribution near the middle portion of its length. The ends will lift up. Part of the beam will be in contact with the support and part will not. Finding the region of contact will involve solving a non-linear contact problem. However, if the same load distribution is doubled in magnitude, the contact region does not change, and all stresses, strains and displacements will double (sticking with linear elasticity for the deformation model of course). The problem is not linear in the mathematical sense, in that the result for two different load distributions can not be superposed. Rather, for a given load distribution "shape", the outcome is proportional to load magnitude. Thus the response is, technically, not linear but homogeneous of degree one in the load (proportional to magnitude but with no superposition in general). Similarly for any two linear elastic conforming bodies that are then loaded somehow and then develop a possibly smaller region of contact.

Large size fasteners are almost never properly tightened. (Required torque is proportional to cube of diameter, with another factor of two or more for high strength.) If they are not properly tightened, they will not perform properly (because low preload is easily overcome, allowing joint slip or fastener fatigue). Using torque to tighten depends strongly on frictional conditions, which are recognized to vary day by day (see Research Council of Structural Connections Specification). The best solution is probably a DTI (direct tension indicator) also called a TIW (tension indicating washer). See Applied Bolting, TurnaSure, and SAE standard J2486.

If a long bar is bolted flat to a half-space with one fastener, and then you lift one end like a prybar, what force does the fastener sustain? In the simple case where the fastener had no preload, it stretches, and the prybar is not forced into contact until some distance away. That distance is determined by prybar stiffness -- the stiffer it is, the straighter it stays, and the further away the point at which it presses the half space and subsequently lies straight. Fastener force is bending moment divided by distance to simple support. Therefore fasteners are far more highly stressed when the prybar is flexible.

How to minimize friction? Normally it is best to reduce shaft diameter, to reduce frictional torque. But the same discussion as above applies. The shaft is loaded at two points, the edge of the hole, and the point where it presses against the bearing bore. If it is flexible, those points will be close together, and the force will be high, leading to friction or even binding.

The greater turning angle occurs at the small sprocket. When a pin link is swinging into or out of engagement, this friction is at a small diameter. When a bushing link is swinging, part of the friction occurs as the exterior of the bushing rubs against its roller (or the sprocket tooth). Oscillating rotation of a pin in a bearing does not necessarily imply sliding -- that depends on friction and clearance and rotation angle. Slip tends to occur after the pin has 'rolled' part way around the bearing bore. So wear does not immediately occur 'on-axis'. With care one could design the joint to allow moderate rolling as manufactured, which should make for lower friction and longer life.

Any number of such springs (even of different k) stretched out and joined at one point yield a composite energy surface which is a sum of circular paraboloids, hence a circular paraboloid. So stiffness is isotropic. Adding a weight to that point is the addition of a planar energy surface, therefore simply shifting the paraboloid. So stiffness is isotropic in the deformed configuration. If a rigid link is pivoted at the weight's equilibrium point, and the spring-supported weight is attached to the link's outer end, its only allowed motion will be at constant potential energy, hence perfectly counterbalanced. Hence, an unbalanced pivoted link can be perfectly counterbalanced with a single zero-length spring grounded above its pivot. Justin Herder in Delft knows a bunch of special things about zero-length springs. The idea is used in the Press long-period seismograph.
A point mass suspended from a zero-length spring follows the 3-D harmonic oscillator equations. The origin is the hanging position. The description remains linear even for large motion. That is, the 3-D pendulum is simpler if the rope is replaced with a zero-length spring.

The difference between the radial force-balance equation (which implies catastrophe above critical speed), and the x and y equations, which imply catastrophe only at resonance (critical speed) if one lingers long enough. A rectangular section shaft with high lateral flexibility is truly dangerous above critical.

A simple homogeneous plate solution is uniform bending, which results in uniform spherical curvature and uniform moments throughout. Test this with a non-convex polygonal shape, with uniform bending moment/length applied all around the boundary.

The 3-D harmonic oscillator is governed by three 1D oscillator equaitons. What if x,y, and z are constrained by a non-holonomic condition on the velocities. Are there any mechanical systems like that? Yes. Using the rolling on Illonator wheels on disks you can make one.

A linear damper provides CV = F = MA Integrating both sides, CL = impulse. Therefore to stop a mass in motion within distance L, choose C = MV/L That is, a damper is defined in terms of impulse / distance. (This is counterintuitive compared to energy -- compressing a damper through a distance L at fixed V takes much more energy, but the impulse is unchanged compared to having the mass come to rest from initial velocity V.)

The normal supposition is that an oscillation decays because energy is destroyed. But it's also possible to pass the energy somewhere else (to be used, if necessary). A simple example is a cart with pivoted, castered wheels, constrained to be parallel (by chain or linkages), and centered by a spring. When lateral oscillation is added to straight-line motion, it damps out. In consequence the vehicle ends up traveling faster. The behavior is something like a sweep oar or a fishes tail. Some bicycles can be propelled by intentionally weaving the steering.

The rigid-bicycle equations of motion, when applied to a conventional geometry and mass distribution, show a particular range of stable speeds. This range is destroyed if the trail or the front wheel spin angular momentum are eliminated. On the other hand the upper speed limit for stability is a gyroscopic phenomenon. It is possible to exhibit a non-gyroscopic (i.e. skates), non-trail design with no-hands stability from a low velocity up to infinity. (Note: typical bicycle instability does not usually bother riders of ordinary skill.). These things are written up here and there in various unpublished documents mostly available on the lab page.

If you tow a two-wheel trailer with an appropriate distribution of mass, and wiggle the tongue laterally, the reaction force it exerts is related to the third time derivative of the displacement -- proportional to velocity and in the same sense. So it is unstable at any speed if connected to a lateral spring and/or mass. Only sufficient damping at the hitch will stabilize it.

Given the equation A*sin(p) + B*cos(p) = C the cleanest solution for p may be written: sin(p) = [C*A - B*sqrt(A^2 + B^2 - C^2)] / sqrt(A^2 + B^2) equally cos(p) = [C*B - A*sqrt(A^2 + B^2 - C^2)] / sqrt(A^2 + B^2). Nothing deep here but several times it has let me (Jim) plow ahead, since ultimately what is wanted is sin(p) or cos(p).

At 12 cents per kw*hr, one watt for one year costs one dollar. Leave a hundred watt bulb on for a year and it costs $100. Run a 1 hp motor all the time, a vacuum cleaner say, and it'll cost you $750.