5.3 Basic Mechanics: Force, Work, Power & Mechanical Energy

5.3 Basic Mechanics: Force, Work, Power & Mechanical Energy

Electricians do not work solely with electrical circuits in isolation; they install, connect, and maintain electromechanical systems including electric motors, winches, passenger lifts, conveyor belts, and air handling fans. Selecting appropriate cable sizes, circuit breakers, and starters for these machines requires understanding how mechanical forces, work done, shaft torque, and power relate directly to electrical demand.

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Mass vs. Weight and Newton's Second Law

Mass vs. Weight

A fundamental physics principle frequently tested in electrotechnical examinations is the distinction between mass and weight:

• Mass ($m$): The intrinsic quantity of matter contained in an object, measured in kilograms ($\text{kg}$). Mass remains constant regardless of position or gravitational field strength.
• Weight ($W$): The downward gravitational force exerted on a mass by Earth's gravity, measured in Newtons ($\text{N}$).

$\text{Weight } (W) = m \times g$ Where:

• $m$ = mass in kilograms ($\text{kg}$)
• $g$ = acceleration due to gravity $= 9.81\text{ m/s}^2$ (standard terrestrial value)

For example, an industrial distribution transformer with a mass of $150\text{ kg}$ exerts a downward gravitational force (weight) of: $W = 150\text{ kg} \times 9.81\text{ m/s}^2 = 1,471.5\text{ N}$

Force and Newton's Second Law of Motion

A force is any interaction that, when unopposed, changes or tends to change the state of rest or uniform motion of an object. Newton's Second Law of Motion states that force is directly proportional to the rate of change of momentum:

$F = m \times a$ Where:

• $F$ = force in Newtons ($\text{N}$)
• $m$ = mass in kilograms ($\text{kg}$)
• $a$ = acceleration in meters per second squared ($\text{m/s}^2$)

Definition of the Newton ($\text{N}$): One Newton is the force required to accelerate a mass of one kilogram at a rate of one meter per second squared ($1\text{ N} = 1\text{ kg}\cdot\text{m/s}^2$).

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Work Done and Mechanical Energy

Work Done ($W$)

In mechanics, work is done whenever a force acts upon an object causing it to move through a distance in the direction of the force:

$W = F \times d$ Where:

• $W$ = work done in Joules ($\text{J}$) or Newton-meters ($\text{N}\cdot\text{m}$)
• $F$ = applied force in Newtons ($\text{N}$)
• $d$ = distance moved in the direction of the force in meters ($\text{m}$)

When a load is lifted vertically against gravity, the force required equals the weight of the load ($F = m \times g$), yielding: $W = m \times g \times h$ Where $h$ is the vertical height lifted in meters.

Mechanical Energy: Potential and Kinetic

Energy is defined as the capacity to do work, measured in Joules ($\text{J}$).

1. Gravitational Potential Energy ($E_p$): The stored energy possessed by a body due to its elevated position in a gravitational field: $E_p = m \times g \times h$
2. Kinetic Energy ($E_k$): The energy possessed by an object due to its motion: $E_k = \frac{1}{2} m v^2$ Where $m$ is mass in $\text{kg}$ and $v$ is velocity in meters per second ($\text{m/s}$).

According to the Principle of Conservation of Energy, energy cannot be created or destroyed; it can only be converted from one form to another. When an electric hoist lowers a heavy cable drum, potential energy converts into kinetic energy and thermal energy dissipated through friction in the brakes.

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Mechanical Power and Torque

Mechanical Power ($P$)

Power is the rate of doing work or the rate of energy transfer over time:

$P = \frac{W}{t} = \frac{F \times d}{t}$ Since linear velocity is distance divided by time ($v = \frac{d}{t}$), power can also be expressed as: $P = F \times v$ Where:

• $P$ = power in Watts ($\text{W}$, where $1\text{ W} = 1\text{ J/s}$)
• $F$ = force in Newtons ($\text{N}$)
• $v$ = constant velocity in meters per second ($\text{m/s}$)

Torque ($\tau$)

Torque (or turning moment) is the measure of the turning effort or rotational force exerted about a pivot axis:

$\tau = F \times r$ Where:

• $\tau$ = torque in Newton-meters ($\text{N}\cdot\text{m}$)
• $F$ = applied perpendicular force in Newtons ($\text{N}$)
• $r$ = radius (perpendicular distance from the axis of rotation to the line of action) in meters ($\text{m}$)

In electric motors, torque is produced by magnetic interactions between the rotor and stator. The rotational mechanical power available at the motor shaft is determined by its torque and rotational speed:

$P = \omega \times \tau = \frac{2\pi N \tau}{60}$ Where:

• $P$ = shaft output power in Watts ($\text{W}$)
• $N$ = rotational speed in revolutions per minute ($\text{RPM}$)
• $\tau$ = shaft torque in Newton-meters ($\text{N}\cdot\text{m}$)
• $\omega$ = angular velocity in radians per second ($\text{rad/s}$), where $\omega = \frac{2\pi N}{60}$

Rearranging to calculate motor shaft torque: $\tau = \frac{60 \times P}{2\pi N}$

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Efficiency and Electro-Mechanical Losses

No electromechanical machine transforms energy with $100\%$ perfection. In every motor, generator, transformer, or gearbox, a portion of the input power is converted into non-useful thermal energy.

┌────────────────────────┐      ┌────────────────────────┐      ┌────────────────────────┐
│      INPUT POWER       │      │   ELECTROMECHANICAL    │      │      USEFUL POWER      │
│         (Pin)          │ ───► │      ENERGY LOSSES     │ ───► │      OUTPUT (Pout)     │
│ Electrical Supply (VI) │      │ • Copper I²R Losses    │      │ Mechanical Shaft Power │
└────────────────────────┘      │ • Iron Core Losses     │      └────────────────────────┘
                                │ • Friction & Windage   │
                                └────────────────────────┘

The Efficiency Formula

Efficiency (symbol $\eta$, Greek letter eta) is the ratio of useful output power to total input power, expressed as a fraction or percentage:

$\eta = \frac{\text{Useful Output Power } (P_{\text{out}})}{\text{Total Input Power } (P_{\text{in}})} \times 100\%$

Rearranging to find the required electrical input power: $P_{\text{in}} = \frac{P_{\text{out}}}{\eta}$

Nature of Losses in Electromechanical Drives

1. Electrical Copper Losses ($I^2 R$): Heat generated in motor stator and rotor copper windings due to electrical resistance.
2. Magnetic Iron Losses: Energy lost in laminated silicon steel cores due to magnetic hysteresis (continuous molecular reorientation) and eddy currents (circulating currents induced in the core metal).
3. Mechanical Losses: Friction in motor bearings, gearbox gear meshes, and aerodynamic resistance (windage) caused by the internal cooling fan.

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Comprehensive Worked Example: Sizing an Electric Hoist Drive

Problem Statement: An electric hoist on a construction site is required to lift a pallet of electrical switchgear with a mass of $600\text{ kg}$ through a vertical height of $18\text{ meters}$ in $24\text{ seconds}$. The hoist gearbox has a mechanical efficiency of $80\%$ ($\eta_m = 0.80$), and the single-phase $230\text{ V}$ electric motor driving it has an electrical efficiency of $75\%$ ($\eta_e = 0.75$). Assuming a motor power factor of $1.0$, calculate:

1. The weight of the load in Newtons.
2. The mechanical work done in Joules.
3. The useful mechanical power output ($P_{\text{out}}$) of the hoist.
4. The total combined system efficiency ($\eta_{\text{total}}$).
5. The electrical power input ($P_{\text{in}}$) drawn from the $230\text{ V}$ supply.
6. The steady-state operating current drawn by the motor.

Step-by-Step Solution:

• Step 1: Weight of the load ($W$): $W = m \times g = 600\text{ kg} \times 9.81\text{ m/s}^2 = 5,886\text{ N}$

• Step 2: Work done ($W$): $W = F \times d = 5,886\text{ N} \times 18\text{ m} = 105,948\text{ J} \approx 105.95\text{ kJ}$

• Step 3: Useful mechanical power output ($P_{\text{out}}$): $P_{\text{out}} = \frac{W}{t} = \frac{105,948\text{ J}}{24\text{ s}} = 4,414.5\text{ W} = 4.415\text{ kW}$

• Step 4: Combined system efficiency ($\eta_{\text{total}}$): $\eta_{\text{total}} = \eta_m \times \eta_e = 0.80 \times 0.75 = 0.60 \text{ (or } 60\%\text{)}$

• Step 5: Electrical power input ($P_{\text{in}}$): $P_{\text{in}} = \frac{P_{\text{out}}}{\eta_{\text{total}}} = \frac{4,414.5\text{ W}}{0.60} = 7,357.5\text{ W} \approx 7.36\text{ kW}$

• Step 6: Motor operating current ($I$): $I = \frac{P_{\text{in}}}{V} = \frac{7,357.5\text{ W}}{230\text{ V}} \approx 31.99\text{ A}$

This calculation demonstrates why mechanics is critical for electricians: the useful mechanical work rate requires only $4.41\text{ kW}$, but due to cumulative losses, the electrical installation must supply $7.36\text{ kW}$ ($32\text{ A}$), dictating a dedicated $32\text{ A}$ or $40\text{ A}$ circuit with suitably rated $6\text{ mm}^2$ or $10\text{ mm}^2$ conductors.