elasticity friction drag and buoyancy models

The mechanical models of elasticity, friction, drag, and buoyancy describe the forces that must be inserted into Newton's second law. For small deformations the elastic force is linear: $F=-kx$, and stress is related to strain by Hooke's law $\sigma=E\varepsilon$; beyond the proportionality limit this model fails. Static dry friction takes values from zero up to $F_{\max}=\mu_s N$, while kinetic friction is approximately constant, $F_k=\mu_k N$, and opposes relative motion. Drag is modeled as linear, $F=-bv$, at low speeds or quadratic, $F=-\frac12 C_D\rho A v^2$, at high speeds; terminal speed occurs when drag balances the driving force. Buoyancy is described by Archimedes' principle, $F_A=\rho_f V_{\text{disp}}g$; a body floats when $F_A=mg$ and sinks when its average density exceeds the fluid density. All these models are idealizations with limits of validity, and they are always used together with Newton's laws.

Direct. A spring of stiffness $k=200\ \mathrm{N/m}$ is stretched by $x=4\ \mathrm{cm}$. Find the magnitude and direction of the elastic force.

Application. A block of mass $m=5\ \mathrm{kg}$ rests on a horizontal table. The friction coefficients are $\mu_s=0.5$ and $\mu_k=0.4$. A horizontal force $F=30\ \mathrm{N}$ is applied to the block. Will the block move, and if so, with what acceleration? Take $g=9.8\ \mathrm{m/s^2}$.

Transfer. A wooden cube of side $a=0.10\ \mathrm{m}$ and density $\rho=600\ \mathrm{kg/m^3}$ floats in water ($\rho_f=1000\ \mathrm{kg/m^3}$). (a) What fraction of the cube's volume is submerged? (b) What minimum mass must be added to the cube so that its top face is level with the water surface? Take $g=9.8\ \mathrm{m/s^2}$.

Safe self-study activity: work in the PhET 'Buoyancy' simulation or an equivalent one, without heating or hazardous substances.

1. Set the fluid density to $1000\ \mathrm{kg/m^3}$ and vary the body density from $200$ to $1200\ \mathrm{kg/m^3}$.
2. For each case, record whether the body floats or sinks, and measure the submerged volume fraction.
3. Check Archimedes' principle: calculate $F_A=\rho_f V_{\text{disp}}g$ and compare it with the body's weight $mg$.
4. Plot the submerged fraction against the ratio $\rho/\rho_f$ and explain its shape.
5. Optionally, if a drag simulation is available, compare the linear and quadratic models and determine the terminal speed.

Show how elastic, friction, drag, and buoyancy forces enter the equations of motion and where the limits of validity of each model lie.

A combined three-part diagram: (1) a graph of Hooke's law $F=-kx$ with the proportionality limit marked; (2) a friction-force graph: as the external force increases, $F_{\text{fr}}$ rises up to $\mu_s N$ and then drops abruptly to $\mu_k N$; (3) a force diagram for a body in a fluid — weight $mg$ downward, buoyant force $F_A$ upward, drag force $F_d$ opposing velocity, plus a graph of $F_d(v)$ with linear and quadratic regions.

A cyclist plus bicycle of total mass $80\ \mathrm{kg}$ rides down a straight $8\%$ descent ($\sin\theta \approx 0.08$). The drag area is $C_d A = 0.40\ \mathrm{m^2}$, air density $\rho = 1.2\ \mathrm{kg/m^3}$, rolling resistance coefficient $\mu_r = 0.005$. At $v = 12\ \mathrm{m/s}$ the air drag is $F_d = \frac{1}{2}\rho C_d A v^2 = 0.5 \cdot 1.2 \cdot 0.40 \cdot 144 = 34.6\ \mathrm{N}$, rolling friction is $F_r = \mu_r mg = 0.005 \cdot 80 \cdot 9.8 = 3.9\ \mathrm{N}$, and the component of weight along the slope is $mg\sin\theta = 80 \cdot 9.8 \cdot 0.08 = 62.7\ \mathrm{N}$. The first two forces together give $38.5\ \mathrm{N}$, so the bicycle still speeds up. A steady speed is reached when $0.24\,v^2 = 62.7 - 3.9$, that is at $v = 15.6\ \mathrm{m/s} \approx 56\ \mathrm{km/h}$; at that speed the air already takes $58\ \mathrm{N}$ out of the $62.7\ \mathrm{N}$ supplied by the descent.

The scenario exercises the quadratic drag law $F_d = \frac{1}{2}\rho C_d A v^2$ with its coefficient $C_d$ and the dry-friction model $f = \mu N$, where $N = mg\cos\theta \approx mg$. Both forces enter the equation of motion along the incline, and the steady speed follows from the equilibrium condition $mg\sin\theta = F_r + F_d$.

A steel hanger cable of a bridge, length $L = 12\ \mathrm{m}$ and cross-sectional area $A = 4.0\ \mathrm{cm^2} = 4.0\times10^{-4}\ \mathrm{m^2}$, is pulled with a force $F = 60\ \mathrm{kN}$; Young's modulus of steel is $E = 200\ \mathrm{GPa}$. The stress is $\sigma = F/A = 6.0\times10^{4} / 4.0\times10^{-4} = 1.5\times10^{8}\ \mathrm{Pa} = 150\ \mathrm{MPa}$, below the yield strength of structural steel, so the deformation is elastic. The strain is $\varepsilon = \sigma/E = 7.5\times10^{-4}$ and the extension is $\Delta L = \varepsilon L = 9.0\ \mathrm{mm}$. The cable stiffness is $k = EA/L = 6.7\times10^{6}\ \mathrm{N/m}$, so an extra load of $1\ \mathrm{kN}$ stretches it by another $0.15\ \mathrm{mm}$ or so.

The scenario exercises linear elasticity: Hooke's law in the form $\sigma = E\varepsilon$ and its equivalent for a whole body, $F = k\Delta L$ with $k = EA/L$. It is precisely this linear stress-strain relation (within the elastic limit) that turns the cable tension into an extension.

A ferry of mass $m = 480\ \mathrm{t} = 4.8\times10^{5}\ \mathrm{kg}$ floats in sea water of density $\rho = 1025\ \mathrm{kg/m^3}$. Its weight is $mg = 4.8\times10^{5} \cdot 9.8 = 4.70\times10^{6}\ \mathrm{N}$, so it displaces a volume $V = m/\rho = 4.8\times10^{5}/1025 = 468\ \mathrm{m^3}$ of water, and the buoyancy force $F_b = \rho V g$ exactly balances that weight. When $40\ \mathrm{t} = 4.0\times10^{4}\ \mathrm{kg}$ of cargo is taken on board, an extra displaced volume of $4.0\times10^{4}/1025 = 39\ \mathrm{m^3}$ is needed; with a waterplane area of about $700\ \mathrm{m^2}$ the ferry settles roughly $39/700 = 0.056\ \mathrm{m} = 5.6\ \mathrm{cm}$ deeper.

The scenario exercises Archimedes' principle: the buoyant force $F_b = \rho_{fluid} V_{disp} g$ equals the weight of the displaced liquid, and the floating condition $\rho_{fluid} V_{disp} g = mg$ links the mass of the vessel to the volume of its submerged part and to its draft.