Deformation and recovery
A can change a body's shape even when equal opposite forces leave its overall zero. A stretched wire, compressed spring or squeezed block is deformed. Elastic deformation disappears when the deforming load is removed within the appropriate range. Plastic deformation leaves a permanent change. The limit of proportionality, elastic limit, yield region and breaking point describe different features of a -deformation or stress-strain curve; they must not all be treated as one point. A material can remain elastic after its response stops being exactly proportional.
For a simple linear spring, the magnitude of applied needed to hold extension x is, with stiffness k in N/m and x in metres. The restoring acts opposite the , so a signed restoring- equation is Fᵣ = −kx. A spring with k = 200 N/m extends 0.05 m under 10 N in its linear range. Extension is the change in length, not the final length. Convert centimetres or millimetres before substituting. A larger k means less extension for the same .
Read the graph before using its
On a graph with vertically and extension horizontally, the linear is k. If the axes are reversed, the is 1/k. A spring balance uses a calibrated relation between load and deformation to indicate . Check the zero reading, units and usable linear range before interpreting a supplied scale. At a yield region, considerable additional deformation may occur with relatively little increase in load. Breaking is physical failure, not merely the first departure from linearity. These are curve interpretations; they are not instructions to overload apparatus.
Stress, strain and Young's modulus
Tensile stress is the stretching divided by the original perpendicular cross-sectional area. Its unit is Pa, equal to N/m². Tensile strain is extension divided by original length and has no unit because it is a ratio of lengths. In the approximately linear elastic range, Young's modulus. It characterises the material response under the stated conditions. Spring stiffness also depends on geometry: for a uniform straight wire in this simple model.
A wire 2.0 m long with area 1.0 mm² carries 20 N. If, convert area to 1.0 × 10⁻⁶ m². Stress is 2.0 × 10⁷ Pa, strain is 1.0 × 10⁻⁴ and extension is 2.0 × 10⁻⁴ m, or 0.20 mm. Do not convert square millimetres using the unsquared length factor. For the same material and , doubling length doubles extension, while doubling area halves it. These comparisons assume the wire remains within the same material-response regime.
is the area under the -extension curve
Stretching a linear spring gradually from zero extension to x requires. The rises during stretching, so final multiplied by the whole extension would overestimate work by a factor of two. For k = 200 N/m and x = 0.05 m, is 0.25 J. More generally, work is the area under the relevant -extension graph. Within the linear elastic model, per original material volume is, measured in J/m³, numerically equivalent to Pa.
Viscosity is internal to relative fluid motion
A fluid's layers can move at different speeds, and viscosity describes the associated to their relative motion. Dynamic viscosity η has SI unit Pa s. Laminar flow has an organised layered pattern; turbulent flow contains irregular fluctuations. Neither word simply means slow or fast without reference to geometry, and viscosity. A Reynolds number compares inertial and viscous effects. For a sphere of diameter 2a moving at v in fluid ρ, one convenient definition is.
Stokes' law gives drag magnitude for a small sphere in creeping flow, meaning Re is much less than one. Here a is radius, not diameter, and v is relative to the fluid. The fluid is treated as Newtonian, the sphere is away from walls, and the motion is sufficiently steady for the model. Do not apply the linear law to every falling object. Large fast objects often have a drag dependence nearer v², and nearby boundaries can alter the .
Terminal comes from balance
A sphere denser than its surrounding fluid accelerates downward initially. As rises, upward drag increases. At terminal the net is zero, so motion continues at constant rather than stopping. Include buoyancy: weight minus upthrust = drag. With sphere ρₛ, fluid ρ𝒻 and radius a, the Stokes-model result is vₜ = 2a²(ρₛ − ρ𝒻)g/(9η). Omitting the difference incorrectly ignores the displaced fluid. A less dense particle can move upward instead, requiring an appropriate direction convention.
For a = 0.50 mm, difference 1,000 kg/m³, and, vₜ ≈ 1.11 × 10⁻³ m/s. If the fluid is, supporting creeping flow in this idealised example. Doubling radius multiplies the predicted terminal by four only while the assumptions remain valid. The equation itself cannot prove that a larger particle still lies in its allowed regime.
Continuity conserves fluid mass
For steady incompressible flow through a pipe without leaks or branches, volume flow rate is constant, so A₁v₁ = A₂v₂. If area falls from 4 cm² to 1 cm² while initial is 2 m/s, final is 8 m/s. Halving pipe radius quarters area and therefore requires four times the at the same flow rate. For compressible flow, mass conservation has the more general form ρ₁A₁v₁ = ρ₂A₂v₂; cannot automatically be cancelled.
Bernoulli relates along an ideal streamline
For steady, incompressible flow with negligible viscous loss and no pump or turbine between the points, p + ½ρv² + ρgh is constant along a streamline. The terms are , kinetic per volume and gravitational potential per volume. At equal heights, a increase from 2 to 8 m/s in fluid of 1,000 kg/m³ corresponds to a decrease of 30,000 Pa. A faster-flow/lower- slogan without these conditions is not a universal rule.
Stokes drag and ideal Bernoulli address different approximations. Strongly viscous dissipation, turbulence, pumps, leaks or significant changes need additional treatment; continuity alone does not the missing relation. Self-check: explain why terminal does not zero , why strain has no unit, and why a narrower pipe does not create fluid. Answers follow from balance, a ratio of lengths and conservation of mass. In every calculation, name the model, convert units and verify that its assumptions fit the situation.