Illustrated Physics
physics, drawn
Series

Free-body — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A block on a 27° incline. Free-body diagram of a block resting on an inclined plane: weight straight down, resolved into a component pressing into the surface and one pulling along it, with friction opposing the slide.

    The slope, and the two directions that make it easy

    An inclined plane looks like a harder problem than a flat one. Split the weight into two components chosen to suit the slope and it becomes an easier one.

    part 1 · mechanics
  2. Which failure comes first. The dividing line between a block that slides and one that tips over, for a horizontal push applied at the top. Sliding needs a force of μ_s times the weight; tipping needs the push's moment about the leading bottom edge to beat the weight's, which is the weight times half the width. The weight appears in both and cancels, so the boundary is the curve aspect ratio = 1/2μ_s and nothing else: not the mass, not how hard the block is pushed, and not what it is made of except through μ. At μ_s = 0.5 the dividing shape is as tall as it is wide; at μ_s = 0.2 it is 2.5 times as tall. Of the 5 objects marked, 3 sit above the line and go over rather than sliding: a paperback, standing, a full filing cabinet, a pint glass.

    Slide or topple

    Push a wardrobe and it goes over; push a brick and it skids. Both are held by the same friction and both are pushed by the same hand, and which of the two failures arrives first has nothing to do with how hard the push is. The floor decides it, by shifting where it pushes back.

    part 2 · mechanics
  3. Four supports, and a whole line of answers. The same top on four supports at the corners of a square, with the load in the same place. Five sets of reactions are drawn and every one of them satisfies all three equilibrium equations exactly — worst residual 1.1e-16 across the whole family — so statics does not prefer any of them. They differ by a multiple of the pattern plus, minus, plus, minus around the square: pressing one diagonal pair harder and the other pair less adds no net force and no moment about either axis, which is exactly what it means for the problem to have a fourth unknown and only three equations. The rigid-body idealisation has not been applied carelessly here; it has been applied correctly, and the answer it gives is that there is no answer. Every member drawn keeps all four reactions positive, so the requirement that a leg can only push narrows the family without closing it. What decides is left out of the model entirely — how much each leg gives under load.

    The table statics cannot settle

    A rigid top on three legs has one possible set of reactions and a rigid top on four has infinitely many, all of them balancing every force and every moment exactly. The extra leg does not make the problem harder; it makes it unanswerable, and the answer has to come from somewhere the model deliberately threw away.

    part 3 · mechanics
  4. The stress a temperature change puts into a bar that cannot move. The stress in a member held between supports that will not let it change length, against how much its temperature changes, for four materials. Each line is EαΔT, computed here from a free expansion and the force needed to undo it, and checked by evaluating it for a bar half a metre long and one thirty-seven metres long: the two agree to every figure carried, because neither the length nor the cross-section appears in the answer. steel develops 2.40 MPa for every kelvin and reaches yield at 104 K; aluminium develops 1.59 MPa for every kelvin and reaches yield at 151 K; concrete develops 0.30 MPa for every kelvin and reaches cracking at 10 K; invar develops 0.17 MPa for every kelvin and reaches yield at 1655 K. Concrete reaches its cracking stress after ten kelvin, which is less than a sunny afternoon, and is why every slab has movement joints in it. Invar is in the comparison because it was made to have a small product: it is as stiff as steel and develops a fourteenth of the stress, which is a statement about the expansion coefficient and nothing else.

    The load nobody applied

    A redundant structure develops forces with nothing on it. Change its temperature and the same extra constraint that made statics unanswerable also refuses the expansion — and a restrained steel member reaches its yield stress after a hundred and four kelvin, a figure that contains no length, no area and no load.

    part 4 · mechanics
  5. Four stress states in one beam, and only one of them has no tension. Stress across the depth of a 300 by 600 millimetre concrete section at the middle of an eight-metre span, compression to the right, for four conditions. Under the load alone the bottom fibre is in tension at 11.1 MPa, which is four times what concrete can carry, so an ordinary reinforced beam cracks there and relies on steel to hold the crack together. With 1,500 kilonewtons of prestress 120 millimetres below the centroid and the full load applied, the section runs from 9.8 to 3.6 MPa and every fibre of it is in compression. The second case is the one that surprises: with the prestress applied and nothing whatever to oppose it, the top fibre is in tension at 1.7 MPa, because a force below the centroid bends the beam upwards. A prestressed beam is at its most vulnerable when nothing is on it, and what rescues it is its own weight: adding that alone brings the top back to 0.3 MPa of compression. Each stress block is checked by integrating it and recovering the force and the moment that produced it.

    A state no load could reach

    The free direction a redundant structure leaves open can be driven on purpose. Tighten a tendon through a concrete beam and its whole stress state moves into the half of the range the material is good at; tighten a bolt hard and the load it carries fluctuates by a fifth of what is applied to it. Both put the structure somewhere no arrangement of external loads could.

    part 5 · mechanics
  6. Four tolerances, four elastic answers, one collapse load. The force in each of a table's four legs against the load on it, for four different manufacturing errors, with the legs made of a material that yields at 25 kilonewtons. While everything is elastic the short diagonal pair takes more than its share by a fixed amount that depends on the error and not at all on the load, so the load at which the first leg reaches its capacity runs from 25 to 100 kilonewtons — a spread of more than a factor of three. Past that point the yielded legs hold a constant force and the others take the rest, and the difference the tolerance made is erased. Every one of the four cases collapses at 100 kilonewtons, which is four times one leg's capacity, checked here to a part in a million across the four. The quantity nobody could compute and the quantity that decides whether the structure stands are not the same quantity.

    The one number the tolerances cannot touch

    A redundant structure's load sharing depends on stiffnesses and manufacturing errors that nobody knows. Its collapse load does not depend on either. Once members yield they hold a known force instead of a force proportional to a displacement, the compatibility equations that needed the unknowns drop out, and the load at which the structure becomes a mechanism follows from a work balance with no stiffness in it at all.

    part 6 · mechanics

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