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Static & kinetic friction, coefficient of limiting friction, viscosity & terminal velocity, Stoke's law — worked examples, calculators, and CBT quiz.
Friction is the resistive force that acts between surfaces in contact and opposes relative motion (or the tendency to move). It is essential to many everyday activities (walking, driving) and engineering applications, but also causes energy losses and wear.
Static friction acts on objects at rest. It adjusts to match the applied tangential force up to a maximum value called the limiting (maximum static) friction Fmax. When the applied force exceeds Fmax, motion begins.
Kinetic (dynamic) friction acts on objects already in motion. Its magnitude is usually slightly less than the maximum static friction.
Scenario: A 10.0 kg box sits on a horizontal floor. You push gently; nothing moves until your push reaches 35 N, after which the box slides and steady motion occurs with a constant friction force of 30 N.
Analysis:
Static friction prevented motion until 35 N. After motion, kinetic friction (30 N) opposes motion. Calculated μl ≈ 0.357 (dimensionless).
Practical note: In many real systems μk < μs. Lubrication or smoother surfaces reduce both values.
The coefficient of limiting friction μl is defined as:
μl = Fmax / N
Where Fmax is the maximum static frictional force measured just before motion starts, and N is the normal force (often equal to weight = mg on a horizontal surface).
Mass = 5.00 kg, Fmax = 30.0 N, g = 9.81 m/s² → N = 49.05 N → μ = 30 / 49.05 = 0.611.
Compute N = mg = 5 × 9.81 = 49.05 N. μ = 30 / 49.05 ≈ 0.611.
Viscosity is a fluid property that measures internal resistance to flow. High-viscosity fluids (honey) flow slowly; low-viscosity fluids (water) flow easily. Viscosity depends on temperature (usually decreases as temperature increases for liquids) and molecular interactions.
A falling object reaches terminal velocity when the downward weight is balanced by upward drag (and buoyant) forces, so net acceleration becomes zero. For small spheres in viscous fluids, Stoke's law gives a good approximation for drag.
Consider a small steel sphere r = 1.0 mm (0.001 m), ρs = 7800 kg/m³, in water (ρf = 1000 kg/m³), η = 0.001 Pa·s, g = 9.81 m/s².
Calculation: v = (2/9) (r² g (ρs - ρf) / η)
r² = 1e-6 m²; (ρs - ρf) = 6800 kg/m³; numerator = 1e-6 * 9.81 * 6800 ≈ 0.066708; divide by η = 0.001 → 66.708; multiply by 2/9 → v ≈ 14.8 m/s.
Note: Stoke's law applies when Reynolds number is very small (laminar flow around sphere). For a steel sphere in water at this size, the Reynolds number may be large and the formula overestimates v. Use Stoke's law only for small particles in very viscous fluids or very small sizes.
Stoke's law for drag on a sphere in a viscous fluid (low Reynolds number) states: Fd = 6 π η r v, where η is dynamic viscosity, r radius, v relative velocity.
At terminal velocity for a falling sphere: weight - buoyancy - drag = 0 ⇒ (4/3) π r³ ρs g - (4/3) π r³ ρf g - 6 π η r v = 0. Solve for v to get:
vt = (2/9) (r² g (ρs - ρf) / η)
A block of mass 2.0 kg rests on an incline. The block just begins to slip when the incline angle is 25°. Find the coefficient of limiting friction.
Solution outline: At impending motion down the plane, component of weight down plane W‖ = mg sin θ. Normal N = mg cos θ. For slipping, limiting friction = mg sin θ. μ = Fmax/N = (mg sin θ)/(mg cos θ) = tan θ. Therefore μ = tan 25° ≈ 0.466.
Compute μ = tan 25° ≈ 0.466 (dimensionless).
Find approximate terminal speed of a small glass bead r = 0.0005 m (0.5 mm), ρs=2500 kg/m³ in oil (ρf=900 kg/m³) with η = 0.1 Pa·s. Use v = (2/9)(r² g Δρ / η).
r² = 2.5e-7; Δρ = 1600; numerator = 2.5e-7 * 9.81 * 1600 ≈ 0.3924e-1? (calculate below)
Compute: r² = 2.5×10⁻⁷; r² g Δρ = 2.5e-7 * 9.81 * 1600 = 0.3924; divide by η = 0.1 gives 3.924; multiply by 2/9 ≈ 0.2222 gives v ≈ 0.873 m/s (approx). Note: check arithmetic carefully; units m/s.
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