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3-Phase Servo AVR (AC Voltage Stabilizer) — Parts, Tests, Repair & Maintenance

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3-Phase Servo AVR (AC Voltage Stabilizer) — Troubleshooting, Repair & Maintenance By Edwin Ogie • December 18, 2025 • -- AC Voltage Stabilizer — 3-phase servo control type (example from user photo) A practical, step-by-step guide to diagnose, repair and maintain 3-phase servo Automatic Voltage Regulators (AVR) / servo voltage stabilizers. Written in simple terms for technicians and maintenance teams working with generators, UPS rooms and factories. Includes videos, spare-parts list, safety checklist, troubleshooting flow and links to internal/external resources. Contents Why this matters In environments with unstable mains (frequent sags, surges or phase imbalance) a servo AVR protects sensitive equipment by continuously adjusting an autotransformer tap via a small servo motor. A well-maintained stabilizer saves equipment, reduces downtime and prevents costly damage. ...

FRICTION

Friction — Edwin Ogie Library (Complete Lesson)

Friction — Complete Lesson (Edwin Ogie Library)

Static & kinetic friction, coefficient of limiting friction, viscosity & terminal velocity, Stoke's law — worked examples, calculators, and CBT quiz.

Introduction & Objectives

Learning objectives

  1. Differentiate between static and dynamic (kinetic) friction.
  2. Determine the coefficient of limiting friction experimentally and by calculation.
  3. Compare the advantages and disadvantages of friction.
  4. Suggest practical ways to reduce friction in machines and everyday life.
  5. Analyze factors affecting viscosity and terminal velocity.
  6. Apply Stoke's law to calculate drag and terminal velocity for small spheres in viscous fluids.

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.

Key idea: Friction depends on the nature of surfaces and the normal force — not directly on contact area for rigid bodies in common cases.

1. Static vs Kinetic (Dynamic) Friction — Explanation & Worked Example

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.

Worked example — Identifying static vs kinetic 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:

  1. Before motion: static friction balances your applied force up to Fmax = 35 N.
  2. Once moving: kinetic friction is 30 N (less than the limiting static friction).
  3. Normal force N = mg = 10 × 9.81 = 98.1 N.
  4. Coefficient of limiting friction μl = Fmax / N = 35 / 98.1 ≈ 0.357.

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.

2. Coefficient of Limiting Friction — How to determine & Example

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).

Experimental setup (simple)

  1. Place a block on a horizontal surface and attach a spring balance horizontally.
  2. Gradually pull with the balance until the block just begins to move — record the balance reading Fmax.
  3. Measure the mass m of the block, compute N = mg, then μ = Fmax/N.

Calculator — find μ

Worked calculation

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.

3. Advantages and Disadvantages of Friction; Ways to Reduce It

Advantages

  • Traction for walking, driving and braking — without friction we would slip.
  • Enables simple machines like belts and pulleys to transmit force.
  • Helps fasten bolts and screws (prevents loosening).
  • Allows control — e.g., using brakes on vehicles.

Disadvantages

  • Energy loss as heat in engines and machines (efficiency loss).
  • Wear and tear of components — requires maintenance.
  • Heat generation can lead to failure in sensitive parts.
  • Requires additional power to overcome in moving systems.

Practical methods to reduce friction

  1. Lubrication: Oils and greases form films separating surfaces and reduce direct contact.
  2. Bearings: Use rollers or ball bearings to convert sliding into rolling friction (much lower).
  3. Surface finish: Polishing and smoother surfaces reduce asperity contact.
  4. Streamlining: For fluids, shape objects to reduce drag (aircraft, cars).
  5. Material selection: Use materials with low μ for sliding applications (e.g., PTFE/Teflon).

4. Viscosity and Terminal Velocity — Explanation & Worked Example

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.

Factors affecting viscosity: temperature, molecular size/shape, pressure (for gases), presence of solutes or suspended particles, and fluid composition.

Terminal velocity (qualitative)

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.

Interactive terminal velocity calculator (Stoke's-law regime)

Worked example (Stoke's-law approximation)

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.

5. Stoke's Law — Derivation sketch & Practical Applications

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)

Applications

  • Estimating settling speeds of small particles in fluids (sedimentation).
  • Design of centrifuges and particle separators (when in the appropriate regime).
  • Cloud physics (fall of water droplets — though corrections often needed).
Reminder: Stoke's law is valid only for laminar (very low Re) flows — check Reynolds number for your case.

Key Terms & Definitions

Static friction
The frictional force that prevents motion up to a maximum value.
Kinetic friction
The friction experienced by moving surfaces; usually lower than static friction.
Coefficient of limiting friction (μl)
Ratio Fmax/N at the point motion starts.
Viscosity
Internal resistance of a fluid to flow (Pa·s).
Terminal velocity
When net force = 0 and acceleration = 0 for a falling object.
Stoke's law
Fd = 6 π η r v for a sphere in viscous fluid (low Re).

6. Worked Examples (More practice)

Example: Block on an incline — find μ

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).

Example: Settling particle in viscous oil (Stoke's law)

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.

7. JAMB-style 30-question CBT Quiz on Friction

Time: 15 minutes. Note: We use a mailto feedback link for convenience — we do not collect personal data automatically. If you want your score recorded, click the email link after submitting. See the privacy note below.

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