1.4 - Materials

Density is mass per unit volume: $\rho=m/V$. In SI units, mass is in kilograms, volume in cubic metres and density in $\mathrm{kg\,m^{-3}}$. The relationship can be rearranged to $m=\rho V$ or $V=m/\rho$.

Identify the material volume represented by the data, convert each measurement before substitution, and check powers of ten carefully. For a regular solid, calculate volume from its dimensions; for an irregular solid, a measured displacement volume may be used.

A sample has mass $0.54\,\mathrm{kg}$ and volume $2.0\times10^{-4}\,\mathrm{m^3}$. Its density is $0.54/(2.0\times10^{-4})=2.7\times10^3\,\mathrm{kg\,m^{-3}}$.

A body in a fluid experiences an upward force because fluid pressure is greater at greater depth. The resulting upthrust equals the weight of fluid displaced: $U=m_{fluid}g=\rho_{fluid}V_{displaced}g$.

A fully submerged object displaces $3.0\times10^{-4}\,\mathrm{m^3}$ of water of density $1000\,\mathrm{kg\,m^{-3}}$. Its upthrust is $(1000)(3.0\times10^{-4})(9.81)=2.9\,\mathrm{N}$.

For a small sphere moving slowly through a fluid with laminar flow, Stokes' law gives viscous drag $F=6\pi\eta rv$. Here $\eta$ is dynamic viscosity in $\mathrm{Pa\,s}$, $r$ is sphere radius and $v$ is speed relative to the fluid. Drag acts opposite to relative motion.

At terminal speed for a falling sphere, $W=U+F_{drag}$. Therefore $F_{drag}=W-U$ can be inserted into Stokes' law to find viscosity or speed. For the liquids considered here, increasing temperature reduces viscosity, so a given sphere reaches a greater terminal speed.

Measure the sphere's diameter in several orientations with a micrometer and average it to obtain $r$. Obtain the sphere's weight and the liquid density, or measure the quantities needed to calculate them. Keep the transparent liquid column vertical and record its temperature.

Place at least three horizontal markers on the tube. Release the sphere centrally without pushing it. Put the first timing marker far enough below the surface for acceleration to have ended. Confirm terminal motion by showing that equal distances take equal times, or that speeds in consecutive marked regions agree.

Measure a marker separation $s$ with a metre rule and time the same point on the sphere crossing the two markers. Repeat and average, then calculate $v=s/t$. Reduce reaction-time error with a longer timed distance or calibrated video, while keeping the sphere in the terminal-speed region.

At terminal speed, use $W=U+6\pi\eta rv$, with $U=\rho_{liquid}V_{sphere}g$, and rearrange for $\eta$. Keep all quantities in SI units and compare repeats before quoting an appropriately precise result.

Hooke's law is $\Delta F=k\Delta x$: change in force is proportional to change in extension or compression while the response remains proportional. The stiffness $k=\Delta F/\Delta x$ is measured in $\mathrm{N\,m^{-1}}$.

Measure extension from the unloaded length, not the total length. On a force-against-extension graph, $k$ is the gradient of the straight proportional section. Rearrange to $\Delta x=\Delta F/k$ when stiffness and force change are known.

A spring length changes from $0.180\,\mathrm{m}$ to $0.230\,\mathrm{m}$ when force increases by $4.0\,\mathrm{N}$. Since $\Delta x=0.050\,\mathrm{m}$, $k=4.0/0.050=80\,\mathrm{N\,m^{-1}}$.

Stress measures force intensity; strain measures fractional length change. Young modulus compares them in the linear elastic region and measures material stiffness: a larger $E$ means more stress is required for the same strain.

Convert diameter to radius and calculate $A=\pi r^2$ for a circular wire. Use extension rather than final length in strain. Equivalently, substituting the definitions gives $E=FL/(A\Delta L)$.

In the initial straight region, gradient $\Delta F/\Delta x$ is stiffness. A curved graph beyond the proportional limit may still include some elastic behaviour, so the limit of proportionality and elastic limit are different ideas.

To decide whether deformation is elastic or plastic, consider what happens after the force is removed. Returning to zero deformation indicates elastic recovery; a non-zero intercept on the deformation axis indicates permanent change.

Plot tensile or compressive stress on the vertical axis against strain on the horizontal axis. Because both quantities account for sample dimensions, the graph describes material response more directly than a force-extension graph.

When comparing samples, use the same feature: gradient for stiffness, breaking stress for resistance to fracture by stress, and strain coordinate for deformation. Read values with axis multipliers and units before calculating.

Clamp a long, thin wire securely and measure its original test length $L$. Measure diameter with a micrometer at several positions and in different orientations, then average and calculate $A=\pi d^2/4$. A long, thin wire gives a larger measurable extension for a given load, reducing percentage uncertainty in extension.

Add known masses gradually, allow the wire to settle, and record extension $\Delta L$ from a fiducial marker or pointer. Calculate tensile force $F=mg$. Repeat readings where practical and keep loads within the linear elastic region so the wire returns to its original length.

Plot stress $F/A$ against strain $\Delta L/L$; the best-fit gradient of the straight region is $E$. Alternatively, plot $F$ against $\Delta L$: if its gradient is $k$, then $E=kL/A$. Use a large gradient triangle and propagate the measured units consistently.

Check micrometer zero error, avoid parallax at the extension scale, and use a tray or shield beneath suspended masses. Do not exceed a safe load or stand below the load. Diameter uncertainty matters strongly because area depends on $d^2$.

A small increase in extension requires work $F\,d x$. Therefore the elastic strain energy change is the area under the force-extension graph between the chosen extensions. Force in newtons multiplied by extension in metres gives joules.

For a linear graph starting at the origin, the area is a triangle: $\Delta E_{el}=\tfrac12F\Delta x$. For example, reaching $F=30\,\mathrm{N}$ at $\Delta x=0.080\,\mathrm{m}$ stores $0.5(30)(0.080)=1.2\,\mathrm{J}$.

For a non-linear graph, divide the region into narrow strips and add trapezium areas, or count grid squares with the axis scale applied. Narrower strips follow the curve more closely. State that the result is an estimate and retain sensible precision.