Applied Physics for Trades

Master the physics principles you need for Canadian apprenticeship exams and real-world applications

1. Electricity Fundamentals

What You Need to Know: Electricity is the flow of electrons through a conductor. Understanding voltage (electrical pressure), current (flow rate), and resistance (opposition to flow) is essential for electrical work, troubleshooting circuits, and calculating power requirements on jobsites. These concepts form the foundation of all electrical systems in Canadian buildings and industrial applications.

Key Concepts

Voltage (V): Electrical potential difference measured in volts. It's the "push" that drives electrons through a circuit.

Current (I): Flow of electrons measured in amperes (amps). It's the "flow" of electricity.

Resistance (R): Opposition to current flow, measured in ohms (Ω). Materials like copper conduct well (low resistance); materials like rubber resist (high resistance).

Ohm's Law
V = I × R

Voltage equals current multiplied by resistance

Power Formulas
P = I × V P = V² ÷ R P = I² × R

Power in watts; choose the formula that matches your known values

Series vs. Parallel Circuits

Series Circuits: Components connected end-to-end. Current is the same through all components. Voltages add up. Total resistance = R₁ + R₂ + R₃

Parallel Circuits: Components connected between the same two points. Voltage is the same across all branches. Currents add up. Total resistance is calculated as: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃

Three-Phase Power

Most industrial and commercial systems use three-phase power, which provides more efficiency than single-phase. Three-phase power delivers constant power (no dips like single-phase) and uses smaller conductors for the same power level.

Three-Phase Power
P = √3 × V × I × Power Factor

√3 ≈ 1.732; power factor typically 0.8-1.0 for industrial loads

Worked Examples

Example 1: Calculate Current Using Ohm's Law

Problem: A 120V circuit has a total resistance of 24 ohms. What is the current?

Step 1: Write Ohm's Law: V = I × R
Step 2: Rearrange to solve for I: I = V ÷ R
Step 3: Substitute values: I = 120V ÷ 24Ω = 5 amps
Answer: The current is 5 amperes
Example 2: Calculate Power

Problem: A motor draws 8 amps at 240V. How many watts is it consuming?

Step 1: Use the formula P = I × V
Step 2: Substitute values: P = 8A × 240V = 1,920 watts
Step 3: Convert to kilowatts: 1,920W ÷ 1,000 = 1.92 kW
Answer: The motor consumes 1,920 watts or 1.92 kW
Example 3: Parallel Circuit Total Resistance

Problem: Three resistors of 12Ω, 12Ω, and 6Ω are connected in parallel. What is the total resistance?

Step 1: Use the parallel formula: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃
Step 2: Substitute: 1/R_total = 1/12 + 1/12 + 1/6 = 0.0833 + 0.0833 + 0.1667 = 0.3333
Step 3: Calculate: R_total = 1 ÷ 0.3333 = 3Ω
Answer: Total resistance is 3 ohms

Try It: Practice Questions

Q1: A 480V industrial circuit has a total resistance of 60Ω. Calculate the current.
Solution: Using Ohm's Law: I = V ÷ R = 480V ÷ 60Ω = 8 amps
Q2: An electric heater draws 15 amps at 240V. How much power does it consume?
Solution: Using P = I × V: P = 15A × 240V = 3,600 watts (3.6 kW)
Q3: What voltage drop will occur across a 10Ω resistance if 3 amps flows through it?
Solution: Using V = I × R: V = 3A × 10Ω = 30 volts

2. Magnetism & Electromagnetism

What You Need to Know: Moving electrical charges create magnetic fields, and changing magnetic fields create electrical currents. This relationship powers motors, generators, transformers, and induction equipment found throughout industrial and construction sites. Understanding transformers especially is critical for managing voltage levels in different parts of electrical systems.

Magnetic Fields & Induction

Every magnet has a magnetic field—an invisible region where magnetic forces act. When a conductor moves through a magnetic field or a magnetic field changes near a conductor, an electrical current is induced in that conductor. This is electromagnetic induction, discovered by Faraday.

The strength of an induced current depends on: the strength of the magnetic field, the speed of change, and the number of wire turns in a coil.

Transformers

Transformers use electromagnetic induction to change voltage levels. A primary coil (input) creates a changing magnetic field, which induces a voltage in a secondary coil (output). The voltage ratio depends on the turns ratio.

Transformer Equation
V_primary / V_secondary = N_primary / N_secondary I_primary / I_secondary = N_secondary / N_primary

As voltage steps down, current steps up (power remains constant, ignoring losses)

AC vs DC

Direct Current (DC): Electrons flow in one direction. Used in batteries, electronics, some industrial equipment. Voltage is constant.

Alternating Current (AC): Electrons reverse direction periodically (50-60 cycles per second = 50-60 Hz). Standard for power grids in Canada (60 Hz). AC is easier to transform to different voltages, so it's ideal for long-distance transmission.

Motor Principles

An electric motor reverses the generator principle. A current-carrying wire in a magnetic field experiences a force (left-hand rule for motors). By switching the current direction using a commutator (DC) or by using AC's alternating nature, the wire continuously rotates.

Worked Examples

Example 1: Transformer Step-Down

Problem: A transformer has 2,400 turns on the primary coil and 200 turns on the secondary coil. The primary voltage is 4,800V. What is the secondary voltage?

Step 1: Use the transformer equation: V_primary / V_secondary = N_primary / N_secondary
Step 2: Rearrange: V_secondary = V_primary × (N_secondary / N_primary)
Step 3: Substitute: V_secondary = 4,800V × (200 / 2,400) = 4,800V × 0.0833 = 400V
Answer: Secondary voltage is 400V
Example 2: Transformer Current Relationship

Problem: A transformer steps down from 480V to 120V. The primary current is 2 amps. What is the secondary current?

Step 1: Calculate turns ratio from voltage: N_primary / N_secondary = 480V / 120V = 4:1
Step 2: Use current formula: I_primary / I_secondary = N_secondary / N_primary = 1 / 4
Step 3: Rearrange: I_secondary = I_primary × (N_primary / N_secondary) = 2A × 4 = 8A
Answer: Secondary current is 8 amps
Example 3: AC Frequency and Period

Problem: Canadian power is 60 Hz AC. What is the period (time for one complete cycle)?

Step 1: Frequency (f) = 60 Hz means 60 cycles per second
Step 2: Period (T) = 1 / f = 1 / 60 seconds
Step 3: T = 0.01667 seconds = 16.67 milliseconds
Answer: One complete AC cycle takes 16.67 milliseconds

Try It: Practice Questions

Q1: A step-up transformer has 500 turns on the primary and 5,000 on the secondary. The input voltage is 120V. What is the output voltage?
Solution: V_secondary = 120V × (5,000 / 500) = 120V × 10 = 1,200V
Q2: An ideal transformer draws 10 amps at 240V on the primary. The turns ratio is 1:5 (primary to secondary). What is the secondary current?
Solution: I_secondary = 10A × 1 / 5 = 2 amps
Q3: At 60 Hz, how many times does the AC current change direction per second?
Solution: At 60 Hz, there are 60 complete cycles per second. Current changes direction twice per cycle: 60 × 2 = 120 times per second

3. Hydraulics & Pneumatics

What You Need to Know: Hydraulic and pneumatic systems use fluids (liquid or compressed gas) to transmit and control power. These systems power heavy equipment, tools, and machinery on construction sites and in factories. Understanding pressure, force, and fluid dynamics is essential for safe equipment operation and system design.

Pascal's Law & Pressure

Pascal's Law states: "Pressure applied to a fluid is transmitted equally in all directions." This principle allows a small force on a small area to create a large force on a large area—the basis of hydraulic lifting and pressing.

Pressure
P = F / A

Pressure (Pa or psi) = Force (N or lbs) ÷ Area (m² or in²)

Hydraulic Force Calculations

In a hydraulic system, the same pressure acts throughout the fluid. Therefore, if you have different-sized pistons/cylinders, you can amplify force:

Hydraulic Force Relationship
F = P × A Mechanical Advantage = A_load / A_effort

Larger output area provides greater force multiplication

Flow Rate

Flow rate tells you how much fluid moves through the system per unit time. It determines speed of operation and power.

Volumetric Flow Rate
Q = V / t Q = A × v

Flow rate (L/min or gal/min) = Volume / Time, or Area × Velocity

Head Pressure (Static Pressure from Fluid Height)

In any fluid system, the weight of the fluid column creates pressure at the bottom. This is important for pump selection and system design.

Head Pressure
P = ρ × g × h

Pressure = Density × Gravity × Height. Or: 1 meter of water ≈ 9.81 kPa; 1 foot ≈ 0.433 psi

Worked Examples

Example 1: Hydraulic Force Multiplication

Problem: A hydraulic system has a pressure of 2,000 psi. The input piston is 1 in² and the output cylinder is 50 in². What force is exerted on the load?

Step 1: Calculate the force on the output: F = P × A
Step 2: Substitute: F = 2,000 psi × 50 in² = 100,000 lbs
Step 3: Convert to tons: 100,000 lbs ÷ 2,000 lbs/ton = 50 tons
Answer: Output force is 100,000 lbs (50 tons)
Example 2: Pressure from Force and Area

Problem: A hydraulic cylinder has a piston area of 20 cm². A force of 40,000 N is applied. What is the pressure in the cylinder?

Step 1: Use the pressure formula: P = F / A
Step 2: Substitute: P = 40,000 N / 20 cm² = 40,000 N / 0.002 m² = 20,000,000 Pa
Step 3: Convert: 20,000,000 Pa = 20 MPa or 2,900 psi
Answer: Pressure is 20 MPa (megapascals)
Example 3: Hydraulic Flow Rate

Problem: A pump delivers 100 liters of fluid per minute into a hydraulic system. If the system line has a cross-sectional area of 5 cm², what is the fluid velocity?

Step 1: Convert flow rate to cm³/s: 100 L/min = 100,000 cm³/min = 1,667 cm³/s
Step 2: Use Q = A × v, rearrange: v = Q / A
Step 3: Substitute: v = 1,667 cm³/s ÷ 5 cm² = 333 cm/s = 3.33 m/s
Answer: Fluid velocity is 3.33 m/s

Try It: Practice Questions

Q1: A hydraulic press has a pump piston with 2 in² area and a load piston with 100 in². If the operator applies 500 lbs of force, what force is exerted on the load?
Solution: First find pressure: P = 500 lbs / 2 in² = 250 psi. Then load force: F = 250 psi × 100 in² = 25,000 lbs
Q2: A water tank is 4 meters tall. What is the pressure at the bottom due to the water column? (Density of water = 1,000 kg/m³)
Solution: P = ρ × g × h = 1,000 kg/m³ × 9.81 m/s² × 4 m = 39,240 Pa ≈ 39.2 kPa or 5.7 psi
Q3: A pneumatic system delivers 50 liters per minute through a 2 cm² line. What is the velocity of the compressed air?
Solution: Convert: 50 L/min = 50,000 cm³/min = 833 cm³/s. v = Q / A = 833 cm³/s / 2 cm² = 416.5 cm/s (4.17 m/s)

4. Thermodynamics & Heat Transfer

What You Need to Know: Heat moves through systems by conduction, convection, and radiation. Energy can be stored as sensible heat (temperature change) or latent heat (phase change like melting or boiling). Refrigeration and HVAC systems rely on controlled heat transfer and the refrigeration cycle. Understanding these principles ensures safe equipment operation and energy-efficient installations.

Sensible vs. Latent Heat

Sensible Heat: Heat that causes a temperature change. When you add sensible heat to water at 20°C, its temperature increases. You can measure this temperature change with a thermometer.

Latent Heat: Heat that causes a phase change (solid → liquid → gas) without changing temperature. When water boils at 100°C, adding more heat doesn't increase temperature; it converts liquid to steam. This "hidden" heat is critical in refrigeration cycles.

Specific Heat Calculation

Different materials require different amounts of heat to raise their temperature by one degree. This is specific heat capacity.

Sensible Heat
Q = m × c × ΔT

Heat (J or BTU) = Mass (kg or lbs) × Specific Heat (J/kg·°C or BTU/lb·°F) × Temperature Change

Heat Transfer Modes

Conduction: Direct transfer of heat through a material. Example: a metal rod heated at one end; heat travels along the rod by molecular vibration.

Convection: Heat transfer through a moving fluid (liquid or gas). Example: warm air rising from a heater spreads heat around a room.

Radiation: Heat transfer via electromagnetic waves. Example: standing near a campfire or feeling the sun's warmth. No medium needed.

BTU and kW Conversions

British Thermal Units (BTU) measure heat in imperial units; kilowatts (kW) in metric. 1 BTU ≈ 1.055 kJ; 1 kW = 3.412 BTU/h

BTU/kW Conversions
Heating Capacity (BTU/h) = Power (kW) × 3,412 Power (kW) = BTU/h ÷ 3,412

HVAC and heating systems often use BTU; electrical systems use kW

Refrigeration Cycle

A refrigerant circulates through four stages: (1) Evaporation (cold side—absorbs heat from the space), (2) Compression (compressor pressurizes the gas), (3) Condensation (hot side—rejects heat to outside), (4) Expansion (pressure drops before returning to evaporator). This cycle continuously moves heat from cold to hot (with energy input).

Worked Examples

Example 1: Sensible Heat Calculation

Problem: You need to heat 50 kg of water from 20°C to 80°C. Specific heat of water = 4,186 J/(kg·°C). How much heat energy is required?

Step 1: Use Q = m × c × ΔT
Step 2: Calculate temperature change: ΔT = 80°C - 20°C = 60°C
Step 3: Substitute: Q = 50 kg × 4,186 J/(kg·°C) × 60°C = 12,558,000 J
Step 4: Convert to kJ: 12,558,000 J ÷ 1,000 = 12,558 kJ
Answer: 12,558 kJ (or about 3.49 kWh) of heat is required
Example 2: BTU to kW Conversion

Problem: An air conditioning unit is rated at 60,000 BTU/h. What is its cooling capacity in kilowatts?

Step 1: Use the conversion formula: kW = BTU/h ÷ 3,412
Step 2: Substitute: kW = 60,000 ÷ 3,412 = 17.58 kW
Answer: The AC unit has a capacity of 17.6 kW
Example 3: Heating Element Power

Problem: A 2 kW electric heater runs for 3 hours. How much heat (in BTU) is produced?

Step 1: Total operating time: 2 kW × 3 hours = 6 kWh
Step 2: Convert to BTU: 6 kWh × 3,412 BTU/kWh = 20,472 BTU
Answer: The heater produces 20,472 BTU of heat

Try It: Practice Questions

Q1: How much heat is needed to raise 25 kg of copper from 15°C to 95°C? (Specific heat of copper = 385 J/(kg·°C))
Solution: Q = 25 × 385 × (95 - 15) = 25 × 385 × 80 = 770,000 J or 770 kJ
Q2: A space heater produces 36,000 BTU/h. What is its power output in kW?
Solution: kW = 36,000 ÷ 3,412 = 10.55 kW
Q3: Which heat transfer method is responsible for the sun warming your skin on a clear day?
Solution: Radiation — electromagnetic waves travel through the vacuum of space, requiring no medium

5. Forces & Mechanics

What You Need to Know: Forces cause motion and deformation. Newton's laws explain how objects move under force, and torque concepts allow you to analyze rotational systems. Mechanical advantage from simple machines (levers, pulleys, gears) lets you do more work with less effort. Rigging and load calculations depend on understanding these principles for safe lifting and moving.

Newton's Laws & Basic Force

Newton's Second Law: Force causes acceleration. The more force applied, the greater the acceleration (if mass is constant). A heavier object requires more force to accelerate at the same rate.

Force, Mass, and Acceleration
F = m × a

Force (N) = Mass (kg) × Acceleration (m/s²); 1 Newton = 1 kg·m/s²

Torque (Rotational Force)

Torque is a twisting or turning force. It depends on both the force applied and the distance from the pivot point. A wrench is long to provide leverage—the same muscle force creates more torque with a longer wrench.

Torque
τ = F × d

Torque (N·m or ft·lbs) = Force (N or lbs) × Perpendicular Distance from Pivot (m or ft)

Mechanical Advantage & Simple Machines

Mechanical Advantage (MA) is the ratio of output force to input force. It tells you how much the machine multiplies your effort.

Lever (Class 1, 2, 3): A rigid bar pivoting on a fulcrum. MA = distance from effort to fulcrum / distance from load to fulcrum

Pulley: A wheel with a rope. Fixed pulleys change direction (MA = 1); movable pulleys multiply force (MA = 2 or more depending on rope arrangement).

Gears: Interlocking wheels with teeth. Gear ratio = Number of teeth on driven gear / Number of teeth on driving gear. Increased teeth ratio increases torque but decreases speed.

Mechanical Advantage
MA = Output Force / Input Force MA = Distance Input Moves / Distance Output Moves

Rule: You gain force but lose distance (or vice versa)

Load Calculations for Rigging

When lifting loads with slings, cables, or equipment, calculate the safe working load (SWL) based on sling angle, number of slings, and material strength. The angle of the sling affects the vertical component of force and the load on each sling.

Sling Tension (Load per Sling)
Tension = Load / (2 × cos(θ))

Where θ is the angle from vertical; as angle increases, tension increases; vertical lift is most efficient (θ = 0°)

Worked Examples

Example 1: Calculate Force from Acceleration

Problem: An 800 kg forklift is accelerating at 0.5 m/s². What force is the engine producing?

Step 1: Use F = m × a
Step 2: Substitute: F = 800 kg × 0.5 m/s² = 400 N
Answer: The engine produces 400 Newtons of force
Example 2: Torque from a Wrench

Problem: A worker applies 50 N of force at the end of a 0.3 m wrench. What torque is applied to a bolt?

Step 1: Use τ = F × d
Step 2: Substitute: τ = 50 N × 0.3 m = 15 N·m
Answer: Torque applied is 15 N·m
Example 3: Lever Mechanical Advantage

Problem: A crowbar has the fulcrum 0.1 m from the load and the effort force applied 0.6 m from the fulcrum. A worker applies 200 N. What is the output force?

Step 1: Calculate MA: MA = 0.6 m / 0.1 m = 6
Step 2: Calculate output force: F_output = F_input × MA = 200 N × 6 = 1,200 N
Answer: Output force is 1,200 N (or about 120 kgf)

Try It: Practice Questions

Q1: A 1,000 kg load hangs from two identical slings at a 60° angle from vertical. What is the tension in each sling?
Solution: Load force = 1,000 kg × 9.81 m/s² = 9,810 N. Tension = 9,810 / (2 × cos(60°)) = 9,810 / (2 × 0.5) = 9,810 N per sling
Q2: A pulley system uses 4 supporting rope segments. What is the mechanical advantage?
Solution: For a movable pulley system, MA = number of supporting ropes = 4. You lift 1 m, the load rises 0.25 m (but with 4× force multiplication)
Q3: A wrench is 0.4 m long. You need to apply 80 N·m of torque to a bolt. What force must you apply?
Solution: F = τ / d = 80 N·m / 0.4 m = 200 N

6. Fluid Mechanics

What You Need to Know: Fluids (liquids and gases) flow through pipes and systems. Understanding flow rate, velocity, pipe sizing, and pressure drop is critical for HVAC systems, plumbing, hydraulics, and pneumatics. Bernoulli's principle explains how pressure and velocity are related in flowing fluids—a foundation for pump selection and system efficiency.

Flow Rate & Velocity

Flow rate (Q) is the volume of fluid moving per unit time. Velocity (v) is how fast the fluid travels. These are related by the pipe's cross-sectional area.

Flow Rate and Velocity
Q = A × v v = Q / A

Flow rate (m³/s or L/s) = Area (m²) × Velocity (m/s)

Continuity Equation

In a closed pipe, the mass of fluid flowing is constant. If the pipe narrows (smaller area), the fluid speeds up. If the pipe widens, the fluid slows down. This is critical for understanding velocity and pressure changes.

Continuity Equation
Q₁ = Q₂ → A₁ × v₁ = A₂ × v₂

If area decreases, velocity increases (and vice versa)

Bernoulli's Principle

For a flowing fluid, total energy is conserved. As the fluid speeds up (high velocity), pressure decreases, and vice versa. This explains why water flows faster in a narrower pipe and why a moving stream of air has lower pressure (demonstrating a vacuum effect).

Bernoulli's Equation
P + ½ρv² + ρgh = Constant

Static pressure + dynamic pressure + hydrostatic pressure = constant

Pressure Drop in Pipes

Friction between the fluid and pipe walls causes pressure to drop along the pipe. Larger pipes and slower flow reduce pressure drop. This affects pump sizing and efficiency.

Pump Head

Pump head is the height of fluid column a pump can lift against gravity. Total pump head includes static head (vertical distance) plus dynamic head (to overcome friction and create flow).

Static vs. Dynamic Pressure

Static Pressure: Pressure due to the fluid's weight and confinement (not moving).

Dynamic Pressure: Pressure due to the fluid's motion. ½ρv²

Total Pressure: Static + Dynamic

Worked Examples

Example 1: Flow Rate and Velocity

Problem: Water flows through a pipe with a cross-sectional area of 0.05 m². The velocity is 2 m/s. What is the flow rate?

Step 1: Use Q = A × v
Step 2: Substitute: Q = 0.05 m² × 2 m/s = 0.1 m³/s
Step 3: Convert: 0.1 m³/s = 100 L/s
Answer: Flow rate is 0.1 m³/s or 100 L/s
Example 2: Continuity Equation - Pipe Narrowing

Problem: Water enters a pipe at point 1 with area 0.1 m² and velocity 1 m/s. At point 2, the pipe narrows to 0.05 m². What is the velocity at point 2?

Step 1: Apply continuity: A₁ × v₁ = A₂ × v₂
Step 2: Calculate: 0.1 m² × 1 m/s = 0.05 m² × v₂
Step 3: Solve for v₂: v₂ = (0.1 × 1) / 0.05 = 2 m/s
Answer: Velocity at the narrower section is 2 m/s
Example 3: Dynamic Pressure

Problem: Air flows through a duct at 10 m/s. The density of air is 1.2 kg/m³. What is the dynamic pressure?

Step 1: Dynamic pressure = ½ρv²
Step 2: Substitute: P_dyn = 0.5 × 1.2 kg/m³ × (10 m/s)² = 0.6 × 100 = 60 Pa
Answer: Dynamic pressure is 60 Pascals

Try It: Practice Questions

Q1: A water line has a flow rate of 50 L/min. The pipe diameter is 2 cm (area = 0.00314 m²). What is the water velocity in the pipe?
Solution: Convert: 50 L/min = 0.000833 m³/s. v = Q / A = 0.000833 / 0.00314 = 0.265 m/s
Q2: A pipe expands from 1 inch diameter (area 5.07 cm²) to 2 inches (area 20.27 cm²). If water enters at 3 m/s, what is the exit velocity?
Solution: v₂ = v₁ × (A₁ / A₂) = 3 m/s × (5.07 / 20.27) = 3 × 0.25 = 0.75 m/s
Q3: Why does water coming from a hose go faster (farther) when you partially block the nozzle?
Solution: Continuity equation — blocking the nozzle reduces the area, so velocity must increase to maintain constant flow rate. Higher velocity means the water travels farther

7. Properties of Materials

What You Need to Know: Different materials respond differently to forces and temperature. Tensile strength tells you how much pulling force a material can withstand before breaking. Hardness indicates resistance to scratching and deformation. Understanding thermal expansion prevents equipment damage when materials heat up. These material properties are critical for selecting the right material for each application in construction, fabrication, and machinery.

Tensile Strength & Yield Point

Tensile Strength: The maximum pulling (tensile) force a material can withstand before breaking. Measured in megapascals (MPa) or psi.

Yield Point: The stress level at which a material starts to permanently deform. Below the yield point, the material springs back; above it, permanent damage occurs.

Steel typically has higher tensile strength than aluminum. Cast iron is strong in compression but weak in tension.

Hardness

Hardness measures resistance to penetration and scratching. Common scales: Rockwell (HRC for hardened steel), Brinell (HB), Mohs (for minerals). Hardness doesn't always correlate with tensile strength; a hard material can be brittle.

Thermal Expansion

Most materials expand when heated and contract when cooled. The coefficient of thermal expansion varies by material. This is why concrete joints and metal expansion loops are installed—to allow for movement without cracking or buckling.

Linear Thermal Expansion
ΔL = L₀ × α × ΔT

Change in length = Original length × Coefficient of thermal expansion × Temperature change

Density

Density is mass per unit volume. It determines how heavy a material is for its size. Used for load calculations, material selection, and understanding buoyancy.

Density
ρ = m / V

Density (kg/m³ or g/cm³) = Mass (kg) / Volume (m³)

Young's Modulus (Stiffness)

Young's modulus measures how much a material resists deformation under stress. High modulus = stiff material (doesn't flex easily). Steel has higher modulus than aluminum.

Common Material Properties (Reference)

Steel: High tensile strength (~400-1000 MPa), good hardness, moderate density (7,850 kg/m³), coefficient of thermal expansion ~12 μm/(m·°C)

Aluminum: Lower tensile strength (~70-400 MPa), lower density (2,700 kg/m³), higher thermal expansion (~23 μm/(m·°C))

Copper: Moderate tensile strength (~200-400 MPa), excellent thermal conductivity, density 8,960 kg/m³

Worked Examples

Example 1: Thermal Expansion of a Steel Beam

Problem: A 10-meter steel beam is installed at 0°C. It will be exposed to temperatures up to 40°C. How much will it expand? (Coefficient of thermal expansion for steel ≈ 12 × 10⁻⁶ /°C)

Step 1: Use ΔL = L₀ × α × ΔT
Step 2: Temperature change: ΔT = 40°C - 0°C = 40°C
Step 3: Substitute: ΔL = 10 m × 12 × 10⁻⁶ /°C × 40°C = 10 × 12 × 40 × 10⁻⁶ = 4,800 × 10⁻⁶ m = 0.0048 m
Answer: The beam expands by 4.8 mm
Example 2: Density Calculation

Problem: A steel plate has a mass of 50 kg and a volume of 0.00636 m³. What is its density?

Step 1: Use ρ = m / V
Step 2: Substitute: ρ = 50 kg / 0.00636 m³ = 7,861 kg/m³
Answer: Density is 7,861 kg/m³ (close to standard steel density of 7,850 kg/m³)
Example 3: Aluminum vs. Steel Weight Comparison

Problem: Compare the weight of a 1 m³ block of steel (density 7,850 kg/m³) versus aluminum (density 2,700 kg/m³).

Step 1: Steel mass: m = ρ × V = 7,850 kg/m³ × 1 m³ = 7,850 kg
Step 2: Aluminum mass: m = 2,700 kg/m³ × 1 m³ = 2,700 kg
Step 3: Ratio: Steel is 7,850 / 2,700 = 2.91 times heavier
Answer: Steel is approximately 2.9 times heavier than aluminum for the same volume

Try It: Practice Questions

Q1: A copper pipe is 5 meters long at 20°C. It will be heated to 80°C. How much will it expand? (Coefficient for copper ≈ 17 × 10⁻⁶ /°C)
Solution: ΔL = 5 × 17 × 10⁻⁶ × (80 - 20) = 5 × 17 × 10⁻⁶ × 60 = 5.1 mm
Q2: An aluminum cube has sides of 0.1 m and a mass of 27 kg. Is it solid aluminum or hollow?
Solution: Volume = 0.1³ = 0.001 m³. Density = 27 kg / 0.001 m³ = 27,000 kg/m³. Aluminum is only 2,700 kg/m³, so it must be hollow or filled with a denser material
Q3: Why do concrete joints need to be left in large slabs and structures?
Solution: Thermal expansion and contraction — as temperature changes, concrete expands and contracts. Joints allow this movement without cracking the structure

8. Work, Energy & Power

What You Need to Know: Work is done when a force moves an object. Energy is the capacity to do work—it exists in many forms (mechanical, thermal, electrical). Power is the rate at which work is done or energy is transferred. Understanding these concepts is essential for calculating equipment performance, estimating energy consumption, and designing efficient systems in all trades.

Work

Work is done when a force causes displacement. If you push with force but nothing moves, no work is done (even though you might be tired!). Work depends on the force component in the direction of motion.

Work
W = F × d × cos(θ)

Work (J or ft·lbs) = Force × Distance × Cosine of angle between force and motion

Kinetic & Potential Energy

Kinetic Energy: Energy of motion. A faster-moving object has more kinetic energy. Doubling speed quadruples kinetic energy (because of the v² term).

Potential Energy: Energy due to position. A raised weight has potential energy. Higher elevation = more potential energy.

Kinetic Energy
KE = ½ × m × v²

Kinetic Energy (J) = Half mass × Velocity squared

Potential Energy (Gravitational)
PE = m × g × h

Potential Energy (J) = Mass × Gravity (9.81 m/s²) × Height

Power

Power is the rate of doing work or transferring energy. A motor that lifts a load quickly is more powerful than one that does it slowly. Power is measured in watts (W), kilowatts (kW), or horsepower (hp).

Power
P = W / t P = F × v

Power (W) = Work (J) / Time (s), or Force × Velocity

Energy Conversions

Energy can convert between forms: electrical → mechanical (motor), mechanical → thermal (friction), thermal → electrical (generator), etc. The total energy is conserved, but some is "lost" as waste heat.

Efficiency

No machine is 100% efficient. Some energy is always lost to friction, heat, and other factors. Efficiency is the ratio of useful output to total input.

Efficiency
Efficiency (%) = (Useful Output / Total Input) × 100%

A typical electric motor is 85-95% efficient; a hydraulic pump is 70-90% efficient

Worked Examples

Example 1: Work Done Lifting a Load

Problem: A crane lifts a 5,000 N load vertically 20 meters. How much work is done?

Step 1: Use W = F × d (force and motion are in same direction, so cos(0°) = 1)
Step 2: Substitute: W = 5,000 N × 20 m = 100,000 J
Step 3: Convert: 100,000 J = 100 kJ
Answer: Work done is 100,000 joules (100 kJ)
Example 2: Power from Work and Time

Problem: A pump does 100,000 J of work in 50 seconds. What is its power output?

Step 1: Use P = W / t
Step 2: Substitute: P = 100,000 J / 50 s = 2,000 W
Step 3: Convert: 2,000 W = 2 kW
Answer: Power is 2,000 watts or 2 kW
Example 3: Kinetic Energy of a Moving Object

Problem: A 1,500 kg vehicle is traveling at 20 m/s. What is its kinetic energy?

Step 1: Use KE = ½ × m × v²
Step 2: Substitute: KE = 0.5 × 1,500 kg × (20 m/s)² = 0.5 × 1,500 × 400 = 300,000 J
Step 3: Convert: 300,000 J = 300 kJ
Answer: Kinetic energy is 300,000 joules

Try It: Practice Questions

Q1: A motor lifting a 2,000 N load takes 40 seconds to raise it 10 meters. What is the motor's power output?
Solution: Work = 2,000 N × 10 m = 20,000 J. Power = 20,000 J / 40 s = 500 watts
Q2: A 10 kg weight is raised 5 meters. What is its gravitational potential energy? (g = 9.81 m/s²)
Solution: PE = 10 kg × 9.81 m/s² × 5 m = 490.5 joules
Q3: A pump consumes 2 kW of electrical power but delivers only 1.6 kW of mechanical power. What is its efficiency?
Solution: Efficiency = (1.6 kW / 2 kW) × 100% = 80%