Four fundamental physical quantities describe the behaviour of mechanical systems: force, torque, speed, and power. Mastery of their definitions, units, interrelationships, and calculations is essential for analysing and designing mechanical systems in VCE Systems Engineering.
KEY TAKEAWAY: Force is the push/pull on an object; torque is the rotational equivalent of force; speed describes how fast motion occurs; power is the rate at which work (energy) is transferred.
Force ($F$) is a push or pull that causes, or tends to cause, a change in motion.
$$F = ma$$
| Symbol | Quantity | SI Unit |
|---|---|---|
| $F$ | Force | Newton (N) |
| $m$ | Mass | kilogram (kg) |
| $a$ | Acceleration | m/s² |
In static mechanical systems (no acceleration), the forces are in equilibrium:
$$\sum F = 0$$
Worked example:
A 20 kg object rests on a surface. What force must a lever exert to lift it?
$$F_{load} = mg = 20 \times 9.8 = 196 \text{ N}$$
EXAM TIP: Always use $g = 9.8$ m/s² (or 10 m/s² if the question specifies) when converting mass to weight force.
Torque ($\tau$) is the rotational equivalent of force — it is the turning effect of a force about a pivot point.
$$\tau = F \times d$$
where:
- $\tau$ = torque (N·m)
- $F$ = force applied (N)
- $d$ = perpendicular distance from the pivot to the line of action of the force (m)
Principle of moments (equilibrium):
$$\tau_{clockwise} = \tau_{anticlockwise}$$
Worked example:
A spanner applies 25 N at 0.3 m from the bolt centre.
$$\tau = 25 \times 0.3 = 7.5 \text{ N·m}$$
If a second force of 15 N acts on the other side at distance $d$:
$\$15 \times d = 7.5 \quad \Rightarrow \quad d = 0.5 \text{ m}$$
VCAA FOCUS: Torque in gear systems: when a gear reduces speed, it multiplies torque proportionally (and vice versa). This is directly related to the gear ratio.
Torque in gear trains:
If gear ratio $GR = T_{driven}/T_{driver}$:
$$\tau_{driven} = \tau_{driver} \times GR \quad (\text{ignoring losses})$$
A 3:1 reduction gear triples the output torque.
In mechanical systems, two types of speed are used:
Linear speed ($v$): Distance per unit time
$$v = \frac{d}{t} \quad \text{(m/s or m/min)}$$
Rotational speed ($N$): Revolutions per minute (rpm) or radians per second
Relationship between linear and rotational speed:
$$v = \omega r = \frac{2\pi N r}{60}$$
where $r$ = radius (m), $N$ = speed in rpm, $\omega$ = angular velocity (rad/s).
Worked example:
A pulley of diameter 0.2 m rotates at 300 rpm. Find the belt speed.
$$r = 0.1 \text{ m}, \quad v = \frac{2\pi \times 300 \times 0.1}{60} = \frac{60\pi}{60} = \pi \approx 3.14 \text{ m/s}$$
Speed in gear trains:
$$\frac{N_{driver}}{N_{driven}} = \frac{T_{driven}}{T_{driver}}$$
A step-up gear train increases output speed; a step-down increases output torque.
REMEMBER: Speed and torque trade off against each other in a gear train. The product (power) is approximately constant (minus losses).
Power ($P$) is the rate at which work is done or energy is transferred:
$$P = \frac{W}{t} = \frac{F \times d}{t} = F \times v$$
For rotational systems:
$$P = \tau \times \omega = \frac{2\pi N \tau}{60}$$
| Symbol | Quantity | SI Unit |
|---|---|---|
| $P$ | Power | Watt (W) |
| $W$ | Work | Joule (J) |
| $t$ | Time | second (s) |
| $\tau$ | Torque | N·m |
| $\omega$ | Angular velocity | rad/s |
Worked example:
A motor produces 8 N·m of torque at 1500 rpm. Find the power output.
$$\omega = \frac{2\pi \times 1500}{60} = 50\pi \approx 157 \text{ rad/s}$$
$$P = 8 \times 157 = 1257 \text{ W} \approx 1.26 \text{ kW}$$
COMMON MISTAKE: Forgetting to convert rpm to rad/s before using $P = \tau\omega$. Always convert: $\omega = 2\pi N/60$.
$$P = F \times v = \tau \times \omega$$
$$\tau = F \times d$$
$$v = \omega \times r$$
These relationships mean that when a gear train reduces speed (smaller $\omega$), torque ($\tau$) increases proportionally to maintain constant power (minus losses).
A machine requires 50 N·m torque at 200 rpm. The motor provides 10 N·m at 1000 rpm.
Required gear ratio:
$$GR = \frac{1000}{200} = 5:1 \text{ reduction}$$
Torque check (ideal):
$$\tau_{out} = 10 \times 5 = 50 \text{ N·m} \checkmark$$
Power check:
$$P_{motor} = \frac{2\pi \times 1000 \times 10}{60} = 1047 \text{ W}$$
$$P_{output} = \frac{2\pi \times 200 \times 50}{60} = 1047 \text{ W} \checkmark \text{ (ideal)}$$
APPLICATION: When selecting a motor for a system, calculate the required torque and speed at the output, then work backwards through the drive train using gear ratios to determine what the motor must provide.
| Quantity | Unit | Conversion |
|---|---|---|
| Force | N | 1 kgf ≈ 9.8 N |
| Torque | N·m | — |
| Linear speed | m/s | 1 km/h = 1/3.6 m/s |
| Rotational speed | rpm or rad/s | $\omega = 2\pi N/60$ |
| Power | W (Watt) | 1 kW = 1000 W |
