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How do simple machines let a small input force move a large load, and at what cost?

Calculate mechanical advantage, velocity ratio and efficiency for simple machines such as levers, pulley systems and inclined planes, and explain the trade-off between force and distance

A QCE Engineering Unit 4 answer on simple machines. Defines mechanical advantage, velocity ratio and efficiency, explains the force-distance trade-off and friction losses, and works through a pulley and a lever with verified arithmetic.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to analyse simple machines (levers, pulley systems, inclined planes, gears, wheel-and-axle) using three linked quantities: mechanical advantage, velocity ratio and efficiency. You need to calculate each, explain why a machine that multiplies force must move the input through a greater distance, and account for the energy lost to friction. This is the quantitative core of the machines half of Unit 4.

The answer

Mechanical advantage

Mechanical advantage (MA) measures how much a machine multiplies the input force:

MA = \frac{\text{load}}{\text{effort}} = \frac{F_L}{F_E}

An MA greater than $1$ means the load is larger than the effort, which is the usual reason for using a machine: a car jack lets a person lift a tonne. MA has no units because it is a ratio of two forces.

Velocity ratio

Velocity ratio (VR), also called the movement ratio, is set purely by the machine's geometry:

VR = \frac{\text{distance moved by effort}}{\text{distance moved by load}} = \frac{d_E}{d_L}

For an ideal (frictionless) machine, energy in equals energy out, so $F_E d_E = F_L d_L$ and the ideal MA equals the VR. Real machines fall short of this.

Efficiency and the trade-off

Efficiency compares the useful work out with the work put in:

\eta = \frac{\text{useful work output}}{\text{work input}} = \frac{MA}{VR}

usually expressed as a percentage. No real machine reaches $100\%$ because some input energy is always lost to friction and, in some machines, to lifting the machine's own moving parts. This is why $MA < VR$ for any real machine.

The force-distance trade-off follows directly from energy conservation. To gain a large mechanical advantage you must accept a large velocity ratio: the effort moves through a much greater distance than the load. A pulley system that halves the effort makes the rope move twice as far. You trade distance for force, never escaping the conservation of energy.

The four types of motion

Machines convert and redirect motion between four types: linear (straight line), rotary (turning about an axis), oscillatory (back and forth through an arc) and reciprocating (back and forth in a straight line). A crank converts rotary motion to reciprocating motion; a rack and pinion converts rotary to linear. Recognising these conversions is part of analysing any mechanism.

A two-pulley lifting system

Find mechanical advantage, velocity ratio and efficiency

A block-and-tackle supports a load of $400\,\text{N}$ using two pulleys, so two rope segments support the load. An effort of $250\,\text{N}$ is needed. To raise the load by $0.50\,\text{m}$ , the effort end of the rope is pulled $1.0\,\text{m}$ . Find the mechanical advantage, velocity ratio and efficiency.

Mechanical advantage:

MA = \frac{F_L}{F_E} = \frac{400}{250} = 1.6

Velocity ratio:

VR = \frac{d_E}{d_L} = \frac{1.0}{0.50} = 2.0

Efficiency:

\eta = \frac{MA}{VR} = \frac{1.6}{2.0} = 0.80 = 80\%

The ideal machine would give $MA = 2.0$ , but friction reduces it to $1.6$ , an efficiency of $80\%$ . Note the trade-off: the effort moves twice as far as the load to halve (ideally) the force required.

Why this matters for machines in society

Every lifting, gripping and driving machine, from a bicycle to a hydraulic excavator, is designed around these ratios. Engineers choose a velocity ratio to get the force multiplication a task needs, then work to raise efficiency by reducing friction with bearings and lubrication. Quoting MA, VR and efficiency together is how you describe a machine's real performance.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 QCAA5 marksExplain the concepts of mechanical advantage and velocity ratio using a simple pulley system. Provide an annotated sketch to support your response.

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Five marks: two for the explanations and three for a correctly annotated sketch (effort, load, and the distances each moves).

Mechanical advantage (MA) is the load divided by the effort, MA = F_L / F_E. In a block-and-tackle pulley system the number of rope sections that support the load sets the MA. For two supporting ropes the effort needed is half the load, so MA = 2:1 [1 mark for wording that links the number of supporting ropes to the proportional reduction in effort].

Velocity ratio (VR) is the distance moved by the effort divided by the distance moved by the load, VR = d_E / d_L. With two supporting ropes the effort end must be pulled through twice the distance the load rises, so VR = 2:1 [1 mark].

Sketch marks: a labelled pulley system showing the effort [1 mark], the load [1 mark], and the distance moved by the effort and by the load [1 mark]. The key idea to convey is that force multiplication is paid for by a proportionally greater effort distance, since efficiency = MA / VR can never exceed 1.

2022 QCAA6 marksA pulley system moves a generator with a mass of 300 kg up a 15 degree incline. The coefficient of static friction between the incline and the generator is 0.3. Determine the tension in the pulley rope required at E to almost begin moving the generator up the incline if the pulley system is 80% efficient. Answer to two decimal places. Assume that the pulley system is parallel to the incline.

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Six marks, one per stage. The pulley system has a velocity ratio of 3 (given by its geometry).

Mechanical advantage from efficiency: MA = efficiency x VR = 0.80 x 3 = 2.4 [1 mark].
Weight of the generator: mg = 300 x 9.8 = 2940 N [1 mark].
Friction force resisting motion up the incline: F_f = mg x mu x cos(theta) = 2940 x 0.3 x cos 15 = 851.95 N [1 mark].
Component of weight along the incline: F_p = mg x sin(theta) = 2940 x sin 15 = 760.93 N [1 mark].
Total resistive load the rope must overcome: F_total = 851.95 + 760.93 = 1612.88 N [1 mark].
The rope tension is the effort, so T = F_total / MA = 1612.88 / 2.4 = 672.03 N [1 mark].

The tension at E required to almost begin moving the generator up the incline is 672.03 N.