Some Mechanical Properties of Matter

When a cylindrical object is under tensile load:

1. Dimensional changes:

   • Length increases from L to L+ΔL
   • Diameter decreases from d to d-Δd
   • Volume changes according to the material's compressibility

2. Physical explanation:

   • Intermolecular bonds along the tensile axis are stretched
   • Conservation of volume (for incompressible materials) forces lateral contraction
   • Molecular rearrangement occurs as atomic bonds reorient toward the direction of applied force

3. Governing relationship:

   • These changes are related by Poisson's ratio: σ = -Δd/d ÷ ΔL/L
   • For most solid materials, σ ranges from 0.2-0.4 (perfect volume conservation would give σ = 0.5)

Example: A rubber band not only becomes longer when stretched, but also noticeably thinner.

Formula for height of liquid rise: $h = \frac{2S\cos\theta}{r\rho g}$

Where:

• h = height of liquid column
• S = surface tension
• θ = contact angle
• r = radius of capillary tube
• ρ = density of liquid
• g = acceleration due to gravity

Key principles:

1. Force balance: Weight of liquid column equals the vertical component of surface tension force
2. Surface tension force: $F_1 = 2\pi rS\cos\theta$ (acting upward)
3. Weight of liquid column: $W = \pi r^2h\rho g$ (acting downward)
4. At equilibrium: $F_1 = W$, giving $2\pi rS\cos\theta = \pi r^2h\rho g$

Important insights:

• Height is inversely proportional to tube radius
• For θ > 90°, cosθ is negative, resulting in depression (negative h)
• For water in glass, θ ≈ 0°, so $h ≈ \frac{2S}{r\rho g}$

Poiseuille's equation: $\frac{V}{t} = \frac{\pi P r^4}{8\eta l}$

Where:

• V/t = volume flow rate
• P = pressure difference between ends
• r = tube radius
• η = fluid viscosity
• l = tube length

Dimensional analysis approach:

1. Assume: $\frac{V}{t} = k\left(\frac{P}{l}\right)^a \eta^b r^c$
2. Dimensional equation: $L^3T^{-1} = (ML^{-1}T^{-2}/L)^a (ML^{-1}T^{-1})^b L^c$
3. Balancing dimensions:
   • For M: 0 = a+b
   • For L: 3 = -2a-b+c
   • For T: -1 = -2a-b
4. Solving: a=1, b=-1, c=4
5. Empirical determination gives k = π/8

Key dependencies:

• Flow rate ∝ r⁴ (extremely sensitive to radius)
• Flow rate ∝ P (linear with pressure difference)
• Flow rate ∝ 1/η (inversely proportional to viscosity)
• Flow rate ∝ 1/l (inversely proportional to tube length)

Solution process:

1. Find the radius of each small droplet:

   • Volume conservation: $1000 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$
   • Where R = 10⁻² m is the original radius and r is the radius of each small droplet
   • This gives: $r = \frac{R}{10} = 10^{-3}$ m

2. Calculate surface areas:

   • Original drop: $A_1 = 4\pi R^2 = 4\pi \cdot (10^{-2})^2 = 4\pi \cdot 10^{-4}$ m²
   • Small droplets: $A_2 = 1000 \cdot 4\pi r^2 = 1000 \cdot 4\pi \cdot (10^{-3})^2 = 40\pi \cdot 10^{-4}$ m²

3. Find increase in surface area:

   • $\Delta A = A_2 - A_1 = 40\pi \cdot 10^{-4} - 4\pi \cdot 10^{-4} = 36\pi \cdot 10^{-4}$ m²

4. Calculate gain in surface energy:

   • $\Delta U = \Delta A \cdot S = 36\pi \cdot 10^{-4} \text{ m}^2 \cdot 0.075 \text{ N/m} = 8.5 \cdot 10^{-4}$ J

Physical significance: This energy must be supplied to break the original drop into smaller droplets. The surface area increases by a factor of 10, demonstrating why atomization requires energy input.

The U-shaped wire frame demonstration illustrates the relationship between surface tension and surface energy through:

• When the sliding wire is pulled outward by distance x:

• The film's surface area increases by 2·l·x (two surfaces)
• Work done against surface tension = F·x = 2·S·l·x
• This work is stored as increased surface energy = S·(2·l·x)

• This shows that surface tension S equals surface energy per unit area:

• Surface energy = work done / area created
• Surface energy/area = (2·S·l·x)/(2·l·x) = S

• The soap film behavior reveals that:

• Surfaces behave like stretched elastic membranes
• They store energy proportional to their area
• They exert tangential forces to minimize their area

This demonstrates why S has equivalent units of N/m (force/length) and J/m² (energy/area).

The excess pressure inside a spherical liquid drop is given by:

$P_2 - P_1 = \frac{2S}{R}$

Where:

• $P_2$ = pressure inside the drop
• $P_1$ = pressure outside the drop
• $S$ = surface tension of the liquid
• $R$ = radius of the drop

This pressure difference arises from surface tension forces acting along the curved surface. When analyzing the equilibrium of a hemispherical portion:

• Surface tension pulls inward along the circular perimeter ($F_1 = 2\pi RS$)
• Pressure forces act on both sides of the hemisphere
• The smaller interior volume requires higher pressure to balance the inward pull of surface tension

This principle explains why smaller droplets have greater internal pressure, as the excess pressure is inversely proportional to radius.

In a soap bubble:

• The pressure inside the bubble (P₂) > pressure in soap solution (P') > atmospheric pressure outside (P₁)
• The excess pressure inside relative to outside is: P₂ - P₁ = 4S/R Where: S = surface tension of the solution, R = radius of the bubble
• This pressure difference arises because:
  1. Each of the two soap film surfaces contributes 2S/R pressure (from surface tension)
  2. The two surfaces combined create 4S/R total pressure difference
• Physical cause: the inward-pulling force of surface tension creates greater pressure on the concave side of any curved surface

Example: A soap bubble with 3 cm radius and solution surface tension of 0.03 N/m has excess pressure of 4(0.03)/0.03 = 4 Pa

When the contact angle equals zero:

1. The liquid molecules adhere perfectly to the capillary wall, causing the meniscus to form a complete hemisphere
2. The volume of this hemispherical meniscus equals $\frac{2\pi r^3}{3}$ (where $r$ is the tube radius)
3. This volume must be accounted for when calculating total liquid rise
4. The weight of liquid in the meniscus is $\frac{1}{3}\pi r^3\rho g$

This precise hemispherical shape occurs because the adhesive forces between liquid and solid completely dominate over the cohesive forces within the liquid at the boundary, making the tangent to the liquid surface perfectly perpendicular to the tube wall.

Dimensional analysis steps:

1. Start with general form: $F = kr^av^b\eta^c$ (where k is dimensionless)

2. Express dimensions on both sides: $[M L T^{-2}] = [L]^a [LT^{-1}]^b [ML^{-1}T^{-1}]^c$

3. Compare exponents of fundamental dimensions: For M: 1 = c For L: 1 = a + b - c For T: -2 = -b - c

4. Solve the system of equations: c = 1 a + b - 1 = 1 -b - 1 = -2

   Therefore: a = 1, b = 1, c = 1

5. Resulting equation: $F = kr\eta v$

6. The dimensionless constant k equals 6π, giving Stokes' law: $F = 6\pi\eta rv$

This derivation assumes laminar flow around a perfect sphere with no-slip boundary conditions.

Potential experimental errors in Stokes' method:

1. Ball size measurement errors:

   • Effect: Radius appears in formula as r²
   • 5% error in radius → ~10% error in viscosity
   • Underestimating radius leads to overestimating viscosity

2. Wall effects:

   • Test tube walls increase drag force
   • Effect: Calculated viscosity appears higher than actual
   • Correction: Use tubes with diameter ≥ 10× ball diameter

3. Non-terminal velocity measurements:

   • If ball hasn't reached terminal velocity before timing
   • Effect: Measured velocity is lower, calculated viscosity higher
   • Prevention: Ensure sufficient distance before timing marks

4. Temperature fluctuations:

   • Viscosity is highly temperature-dependent
   • Effect: Inconsistent results between trials
   • Control: Maintain constant temperature with water bath

5. Density measurement errors:

   • Both ball and fluid density affect calculations
   • Effect: Directly proportional errors in calculated viscosity
   • Importance: Critical for accurate results

6. Timing errors:

   • Human reaction time in stopwatch operation
   • Effect: Significant for fast-moving objects
   • Improvement: Use electronic timing or video analysis