Night-before formula sheet
Every formula, ratio shortcut and trap for Solids and Fluids on one page. Read it once tonight and once in the morning.
CHAPTER 8
Mechanical Properties of Solids
Stress\( \sigma = \dfrac{F}{A} \)N m⁻² = Pa, dimensions [ML⁻¹T⁻²]. Internal restoring force per area.Symbols, assumptions and explanation →
Strains\( \dfrac{\Delta L}{L},\quad \theta \approx \dfrac{\Delta x}{L},\quad \dfrac{\Delta V}{V} \)Longitudinal, shear, volumetric. Dimensionless.Symbols, assumptions and explanation →
Hooke's law\( \text{stress} = E \times \text{strain} \)Valid only up to the proportional limit. Modulus has stress's units.Symbols, assumptions and explanation →
Young's modulus\( Y = \dfrac{FL}{A\,\Delta L} \Rightarrow \Delta L = \dfrac{FL}{\pi r^2 Y} \)Solids only. Steel ≈ 2×10¹¹ Pa. Rigid body: Y = ∞.Symbols, assumptions and explanation →
Shear modulus\( G = \dfrac{F/A}{\theta} \)Solids only. Usually G ≈ Y/3.Symbols, assumptions and explanation →
Bulk modulus\( B = -\dfrac{p}{\Delta V/V},\quad k = \dfrac1B \)Solids, liquids, gases. Water 2.2×10⁹ Pa. Fractional change in radius = p/3B.Symbols, assumptions and explanation →
Poisson's ratio\( \sigma = -\dfrac{\Delta d/d}{\Delta L/L} \)Most conventional materials: 0 to 0.5; auxetic materials can have negative ratios. \( Y = 2G(1+\sigma) = 3B(1-2\sigma) \).Symbols, assumptions and explanation →
Elastic energy\( U = \tfrac12 F\Delta L = \dfrac{YA(\Delta L)^2}{2L} \)Per volume: \( u = \tfrac12\,\text{stress}\times\text{strain} = \dfrac{\text{stress}^2}{2Y} \). Mass M stretching wire by l: U = Mgl/2.Symbols, assumptions and explanation →
Wire as a spring\( k = \dfrac{YA}{L} \)Series: \( 1/k = 1/k_1 + 1/k_2 \). Parallel: \( k = k_1 + k_2 \).Symbols, assumptions and explanation →
Own weight\( \Delta L = \dfrac{MgL}{2AY} = \dfrac{\rho g L^2}{2Y} \)Half of what the same weight hung at the end would produce.Symbols, assumptions and explanation →
Thermal stress\( F = YA\alpha\,\Delta T \)Rod clamped between rigid walls. Independent of length.Symbols, assumptions and explanation →
Beam bending\( \delta = \dfrac{W\ell^3}{4bd^3Y} \)Doubling depth d cuts sag 8×, which is why I-beams are deep.Symbols, assumptions and explanation →
Symbol guide: F and W are force magnitudes in N; M is mass in kg; A is cross-sectional area in m²; L or ℓ is length in m; Y, G and B are moduli in Pa. σ denotes stress in the stress card and Poisson's ratio in the Poisson card. Δ means a signed change; p in the compression formulas means the added pressure. In beam bending, b is breadth, d is vertical depth and δ is sag.
Ratio shortcuts
| Situation | Use |
|---|---|
| Same material, same force, different L and d | \( \Delta L \propto L/d^2 \) |
| Same material, same volume (wire redrawn) | \( \Delta L \propto L^2 \propto 1/A^2 \) |
| Same extension needed, areas A and 3A, same volume | F ratio = 9 (since \( F \propto A^2 \)) |
| Breaking load | \( \propto \) area. Breaking stress is a material constant. |
| Y, G, B, breaking stress | Depend only on the material (and temperature), never on L or A |
CHAPTER 9 · PART I
Fluids at rest
Pressure\( P = \dfrac{F}{A} \)Scalar. Acts equally in all directions at a point; force on a surface is normal to it.Symbols, assumptions and explanation →
Depth\( P = P_0 + \rho g h \)Same depth, same pressure, whatever the vessel's shape (hydrostatic paradox).Symbols, assumptions and explanation →
Gauge pressure\( P - P_0 = \rho g h \)1 atm = 1.013×10⁵ Pa = 76 cm Hg. 1 bar = 10⁵ Pa. 1 torr = 1 mm Hg.Symbols, assumptions and explanation →
U-tube, two liquids\( \rho_1 h_1 = \rho_2 h_2 \)Measured from the common interface level.Symbols, assumptions and explanation →
Pascal / hydraulic lift\( \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \)\( F_2 = F_1 (d_2/d_1)^2 \). Distances: \( A_1x_1 = A_2x_2 \). Work in = work out.Symbols, assumptions and explanation →
Buoyancy\( F_B = \rho_\ell V_\text{sub}\, g \)Equals weight of liquid displaced, floating or sunk.Symbols, assumptions and explanation →
Floating\( \dfrac{V_\text{sub}}{V} = \dfrac{\rho_\text{body}}{\rho_\ell} \)Interface of two liquids: \( \rho V = \rho_1V_1 + \rho_2V_2 \).Symbols, assumptions and explanation →
Apparent weight\( W' = W\left(1 - \dfrac{\rho_\ell}{\rho_\text{body}}\right) \)Loss of weight = buoyant force.Symbols, assumptions and explanation →
Connected cylinders\( h_f = \dfrac{A_1h_1+A_2h_2}{A_1+A_2} \)A₁, A₂ = vessel areas; h₁, h₂ = initial depths above one bottom level; h_f = final depth. Equal surface pressures and same liquid.Symbols, assumptions and explanation →
Gravity's work: equal-area cylinders\( W_g = \dfrac{\rho g A}{4}(h_1-h_2)^2 \)W_g in J; ρ = liquid density; A = area of each vessel; g = gravity. Gravitational potential energy lost during equalisation.Symbols, assumptions and explanation →
Accelerating container\( \tan\theta=|a|/g,\quad |\Delta h|=|a|\ell/g \)a = horizontal acceleration, ℓ = tank length, θ = tilt, Δh = level difference. Vertical lift: g_eff = g + a, with a positive upward.Symbols, assumptions and explanation →
CHAPTER 9 · PART I
Fluids in motion
Continuity (mass)\( A_1v_1 = A_2v_2 \)\( v \propto 1/d^2 \). Branches: \( A v = \sum a_i v_i \).Symbols, assumptions and explanation →
Bernoulli (energy)\( P + \tfrac12\rho v^2 + \rho g h = \text{const} \)P = pressure (Pa), ρ = density (kg/m³), v = speed (m/s), g = gravitational acceleration (m/s²), h = upward elevation (m). Along an ideal steady-flow streamline; at equal elevation, faster flow corresponds to lower pressure.Symbols, assumptions and explanation →
Torricelli\( v = \sqrt{2gh} \)h = depth below the free surface. With extra surface pressure ΔP: \( v^2 = 2gh + 2\Delta P/\rho \).Symbols, assumptions and explanation →
Range of jet\( R = 2\sqrt{h(H-h)} \)Max = H at h = H/2. Depths h and H−h give equal ranges.Symbols, assumptions and explanation →
Time to empty\( t = \dfrac{A}{a}\sqrt{\dfrac{2H}{g}} \)\( t \propto \sqrt{H} \).Symbols, assumptions and explanation →
Venturimeter\( \rho g \Delta h = \tfrac12\rho(v_2^2 - v_1^2) \)With \( A_1v_1 = A_2v_2 \): \( v_1 = \sqrt{\dfrac{2g\Delta h}{(A_1/A_2)^2 - 1}} \).Symbols, assumptions and explanation →
Lift\( F = \tfrac12\rho(v_\text{top}^2 - v_\text{bottom}^2)\,A \)Small difference: \( \Delta v/v = \Delta P/\rho v^2 \). Roof: \( F = \tfrac12\rho v^2 A \), upward.Symbols, assumptions and explanation →
Jet thrust\( F = \rho a v^2 \)Two holes at depth difference h on opposite walls: net 2ρagh.Symbols, assumptions and explanation →
Stagnation rise\( h = \dfrac{v^2}{2g} \)Pitot-type tube facing the flow.Symbols, assumptions and explanation →
CHAPTER 9 · PART II
Viscosity & terminal velocity
Newton's law\( F = \eta A \dfrac{dv}{dx} \)η in Pa·s; 1 poise = 0.1 Pa·s. [ML⁻¹T⁻¹]. Ideal fluid: η = 0.Symbols, assumptions and explanation →
Terminal velocity\( v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta} \)∝ r² and ∝ (ρ − σ). Rising bubble: \( v = 2r^2\sigma g/9\eta \).Symbols, assumptions and explanation →
n drops merge\( v' = n^{2/3} v \)8 drops → 4v. One sphere broken into n: each gets \( v/n^{2/3} \).Symbols, assumptions and explanation →
At terminal velocity\( F_\text{viscous} = W - F_B \)If liquid density = ½ body density: viscous force = Mg/2.Symbols, assumptions and explanation →
Reynolds number\( R_e = \dfrac{\rho v d}{\eta} \)ρ = fluid density, v = mean speed, d = pipe diameter, η = viscosity. Your coaching convention: < 1000 laminar, > 2000 turbulent; transition depends on the setup.Symbols, assumptions and explanation →
Viscous pipe flow\( Q=\dfrac{\pi r^4\Delta P}{8\eta\ell},\quad v_\text{max}=2\bar v \)Q = volume flow rate; r = inner radius; ΔP = pressure drop; η = viscosity; ℓ = pipe length; v̄ = mean speed. Fully developed laminar flow in a circular pipe with no slip.Symbols, assumptions and explanation →
| When temperature rises | Liquids | Gases |
|---|---|---|
| Viscosity η | decreases | increases |
| Surface tension T | decreases, 0 at critical temperature | — |
CHAPTER 9 · PART II
Surface tension & capillarity
Surface tension\( T = \dfrac{F}{L} = \dfrac{W}{\Delta A} \)N/m = J/m². Scalar. Salt raises it, soap lowers it.Symbols, assumptions and explanation →
Film on a frame\( F = 2TL \)Two surfaces. Square frame: 8TL. Ring/disc with hole: 2π(R+r)T.Symbols, assumptions and explanation →
Work on film / bubble\( W = 2T\Delta A,\ \ W_\text{bubble} = 8\pi T(r_2^2 - r_1^2) \)Drop (one surface): \( 4\pi T(r_2^2 - r_1^2) \).Symbols, assumptions and explanation →
Drops merging\( E = 4\pi r^2T\,(n - n^{2/3}) \)Energy released (area falls). Splitting needs energy.Symbols, assumptions and explanation →
Excess pressure\( \text{drop } \dfrac{2T}{R},\ \text{bubble } \dfrac{4T}{R},\ \text{air bubble } \dfrac{2T}{R} \)Air bubble at depth h: \( P_\text{in} = P_0 + \rho g h + 2T/r \).Symbols, assumptions and explanation →
Two bubbles joined\( r = \dfrac{r_1 r_2}{r_2 - r_1} \)Interface radius. Small bubble shrinks into the big one.Symbols, assumptions and explanation →
Capillary rise\( h = \dfrac{2T\cos\theta}{r\rho g} \)r = inner radius. hr = const. Mass raised ∝ r: \( mg = 2\pi rT\cos\theta \).Symbols, assumptions and explanation →
Capillary experiment\( T\approx\dfrac{\rho grh}{2},\quad \dfrac{\Delta T}{T}\approx\dfrac{\Delta r}{r}+\dfrac{\Delta h}{h} \)T = surface tension; ρ = liquid density; r = inner radius; h = vertical rise; g = gravity. θ ≈ 0; neglect meniscus correction. Error sum is a first-order maximum estimate with ρ and g exact.Symbols, assumptions and explanation →
Tilted tube\( \ell = \dfrac{h}{\cos\alpha} \)Vertical height same; α measured from the vertical.Symbols, assumptions and explanation →
Angle of contact\( \theta < 90^\circ \) wets, rises, concaveθ > 90°: doesn't wet, falls, convex (mercury ≈ 140°). Water–glass ≈ 0°.Symbols, assumptions and explanation →
FROM YOUR ✗ MARKS
Traps you have already met
Area ∝ d². Hydraulic press 1.4 cm and 7 cm: ratio 25, not 5 (Test-05 Q5).
Pressure has no direction. Only the force it produces does (Test-05 Q3).
Coin falls off a block: block rises, water level falls (Test-05 Q12).
Variable density: integrate for the average; don't take the middle value (Test-05 Q13).
Buoyancy in an ordinary hydrostatic liquid equals the displaced liquid’s weight; the gravity-driven contribution vanishes in ideal free fall (Test-05 Q15).
Steady flow means the velocity at a point is fixed, not each particle's velocity (Test-05 Q16).
Bernoulli: P + ρgh + ½ρv². Only the KE term has ½. It is energy conservation (Q21, Q42).
Max jet range is at h = H/2 and equals H (Test-05 Q30).
Drops merging release energy, not mass → energy (Test-06 Q7).
Capillary rise depends on the inner radius only (Test-06 Q16).
Terminal velocity ∝ r², not r (Test-06 Q24).
Count every edge: disc with hole 2π(R+r)T, frame 8lT (Test-06 Q25).
Surface tension is a scalar (Test-06 Q27).
Lubricant needs enough viscosity to maintain its film (Test-06 Q43); operating conditions determine the practical choice.