How is mechanical advantage of a pulley system calculated?

Mechanical advantage is the ratio of load force to effort force. For an ideal (frictionless) pulley system the ideal mechanical advantage equals the number of rope segments supporting the moving load block, and this also equals the velocity ratio. To find the real (actual) mechanical advantage, multiply the velocity ratio by the efficiency: AMA equals efficiency times VR. The effort force is then the load divided by the actual mechanical advantage.

How do gear trains multiply torque?

For gear stages connected in series the overall gear ratio is the product of the individual stage ratios. Torque is multiplied by the gear ratio and reduced by efficiency, while output speed is the input speed divided by the gear ratio. So a 20:1 reduction multiplies torque about twentyfold (less losses) and divides speed by twenty. Power out equals power in times overall efficiency.

What is a factor of safety and how is the safe working load found?

The factor of safety is the ratio of a component's breaking (failure) load to the maximum load it is permitted to carry in service. The safe working load equals the minimum breaking load divided by the factor of safety. Australian Standard AS1418 sets the factor of safety by duty: typically 5 for powered crane hoist ropes and 12 for passenger lifts, the higher values allowing for shock loading and wear between inspections.

How does hydraulic lifting use Pascal's principle?

Pascal's principle states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid. In a hydraulic jack a small input force on a small piston creates a pressure that acts on a large output piston, so the output force is larger in the ratio of the piston areas. Because the fluid volume is conserved, the large piston moves a smaller distance, so force is multiplied at the cost of distance, exactly like a pulley.

Why do hoists use worm gears?

A worm gear pair gives a high reduction ratio in a single stage and can be made self-locking, meaning the worm can drive the gear but the gear cannot back-drive the worm. The suspended load is therefore held in place even if power is lost. The trade-off is lower efficiency (heat loss). Australian Standard AS1418 still requires a separate mechanical brake; self-locking worm gearing alone is not accepted as the only holding device.

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NSWEngineering Studies

HSC Engineering Studies Lifting Devices: deep-dive 2026 guide

Deep-dive on the HSC Engineering Studies Lifting Devices module. Mechanical advantage and velocity ratio in pulley systems, efficiency, gear-train torque multiplication, motor sizing, hydraulic lifting, wire rope and factors of safety, with worked calculations and exam-style questions.

Generated by Claude Opus 4.716 min readNESA-ENG-LIFTINGUpdated 2026-05-24

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How Lifting Devices fits into HSC Engineering Studies
Mechanical advantage and velocity ratio
Gear trains and torque
Hydraulic lifting and Pascal's principle
Wire rope and factor of safety
Common Lifting Devices examiner traps
Check your knowledge

How Lifting Devices fits into HSC Engineering Studies

Lifting Devices is the module about machines that trade force for distance. A crane, a hoist, a block and tackle and a hydraulic jack all let a small input move a large load, and in every case the central idea is the same: energy is conserved, so if you reduce the force by a factor you must move the input that same factor further (plus whatever is lost to friction). The calculations divide into four families: pulley mechanical advantage, gear-train torque multiplication, hydraulic pressure, and wire-rope factors of safety.

NESA examines this module quantitatively, often with multi-part questions that build from an ideal calculation to a real one with efficiency, then to a motor-sizing or safe-working-load conclusion. The Australian context (tower cranes on Sydney CBD sites, ship-to-shore container cranes at Port Botany, mining draglines) supplies the named examples.

Mechanical advantage and velocity ratio

For an ideal (frictionless) pulley system, the ideal mechanical advantage equals the velocity ratio, and both equal the number of rope segments supporting the moving load block:

The single most important rule: the mechanical advantage equals the number of rope segments supporting the moving load block, not the total number of pulleys and not the segments at the fixed anchor block.

Configuration	$n$ (segments)	IMA	VR
Single fixed pulley	1	1	1
Single movable pulley	2	2	2
2-and-1 block and tackle	3	3	3
2-and-2 block and tackle	4	4	4
3-and-2 block and tackle	5	5	5

Worked example: block and tackle with efficiency

A block-and-tackle system has four rope segments supporting the load block. It lifts a 2400 N load at an efficiency of 80 percent. Find (a) the ideal mechanical advantage, (b) the velocity ratio, (c) the actual effort force, and (d) the length of rope the operator must pull to raise the load by 1.5 m.

(a) Ideal mechanical advantage: Equal to the number of supporting segments: $IMA = n = 4$ .
(b) Velocity ratio: For an ideal pulley system $VR = n = 4$ .
(c) Actual effort force: First the actual mechanical advantage:

AMA = \eta \times VR = 0.80 \times 4 = 3.2

F_{\text{effort}} = \frac{F_{\text{load}}}{AMA} = \frac{2400}{3.2} = 750 \text{ N}

(d) Input distance. The operator pulls a length equal to the load distance times the velocity ratio:

d_{\text{effort}} = d_{\text{load}} \times VR = 1.5 \times 4 = 6.0 \text{ m}

Check the energy: input work $= 750 \times 6.0 = 4500$ J, output work $= 2400 \times 1.5 = 3600$ J, efficiency $= 3600/4500 = 0.80$ . Consistent. The ideal effort would have been $2400/4 = 600$ N; friction raises it to 750 N.

Real pulleys lose efficiency to bearing friction and rope bending stiffness, and the loss grows as more pulleys are added. This is why most cranes use no more than about six rope falls; beyond that, the extra friction outweighs the further force reduction.

Gear trains and torque

A gear train changes speed and torque. For stages connected in series the overall gear ratio is the product of the stage ratios (never the sum), torque is multiplied and speed is divided.

Torque from a drum and the rope force it produces are linked by the drum radius: $T = F r$ , so $F = T / r$ . The lifting speed of the rope is the drum's angular velocity times its radius: $v = \omega r$ . Converting motor speed from rpm to rad/s uses $\omega = \text{rpm} \times 2\pi / 60$ .

Worked example: hoist load and lifting speed

A hoist motor produces 50 N m of torque at 1450 rpm. It drives a winch drum of radius 0.20 m through a two-stage gear reduction (first stage 4:1, second stage 5:1) at 85 percent overall efficiency. Find the maximum load it can lift and the lifting speed.

Overall gear ratio: $GR = 4 \times 5 = 20$ .

Ideal drum torque: $T_{\text{drum,ideal}} = T_{\text{motor}} \times GR = 50 \times 20 = 1000$ N m.

Actual drum torque (after efficiency): $T_{\text{drum}} = \eta \times T_{\text{drum,ideal}} = 0.85 \times 1000 = 850$ N m.

Maximum load (force on the rope at the drum):

F_{\text{load}} = \frac{T_{\text{drum}}}{r} = \frac{850}{0.20} = 4250 \text{ N}

Drum speed: $\omega_{\text{drum,rpm}} = 1450 / 20 = 72.5$ rpm. In rad/s: $\omega = 72.5 \times 2\pi / 60 = 7.59$ rad/s.

Lifting speed:

v = \omega r = 7.59 \times 0.20 = 1.52 \text{ m/s}

Power check: input $P_{\text{in}} = T\omega = 50 \times (1450 \times 2\pi / 60) = 7.59$ kW; output $P_{\text{out}} = F v = 4250 \times 1.52 = 6.46$ kW; ratio $6.46/7.59 = 0.85$ , matching the stated efficiency.

A worm gear pair (a screw meshing with a gear wheel) gives a high reduction in one stage and can be self-locking: the worm drives the gear but the gear cannot back-drive the worm, so the load is held when power is removed. The cost is lower efficiency (around 40 to 70 percent). Self-locking is a geometry property, not a substitute for a brake; AS1418 still requires a separate mechanical brake on a hoist.

Hydraulic lifting and Pascal's principle

A hydraulic jack multiplies force using a confined fluid. Pascal's principle says the pressure applied to a confined fluid is transmitted undiminished, so the pressure is the same on both pistons.

Worked example: hydraulic jack force multiplication

A hydraulic jack has an input piston of diameter 20 mm and an output piston of diameter 100 mm. An input force of 200 N is applied. Find the output lifting force and the distance the load rises when the input piston is pushed down 60 mm. Assume the system is ideal.

Piston areas (the ratio is what matters, and it equals the diameter ratio squared):

\frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{100}{20}\right)^2 = 5^2 = 25

Output force:

F_2 = F_1 \frac{A_2}{A_1} = 200 \times 25 = 5000 \text{ N}

Output distance (volume conserved): $d_2 = d_1 \dfrac{A_1}{A_2} = 60 \times \dfrac{1}{25} = 2.4$ mm.

The jack multiplies force 25 times but the load rises only 2.4 mm for 60 mm of input travel. Work in equals work out for the ideal case: $200 \times 0.060 = 12$ J, and $5000 \times 0.0024 = 12$ J.

Wire rope and factor of safety

Cranes and hoists run on steel wire rope built from high-tensile wires laid into strands around a fibre core (FC) or independent wire rope core (IWRC). The rope is rated by its minimum breaking load (MBL), the load at which a new rope fails. The load it is allowed to carry in service is the safe working load (SWL).

The high factor for personnel lifts allows for shock loading, dynamic snatch loads and wear between scheduled inspections. Ropes are retired from service when broken-wire counts, diameter reduction (7 percent or more), corrosion or deformation exceed the AS2759 thresholds.

Worked example: safe working load with reeving

A tower crane hoist uses 6 x 36 IWRC wire rope with a minimum breaking load of 215 kN. AS1418 requires a factor of safety of 5 for crane hoist ropes. Find the safe working load of a single rope, and the maximum hook load if the lift uses a 4-fall reeving.

Single-rope SWL:

SWL = \frac{MBL}{FoS} = \frac{215}{5} = 43 \text{ kN}

Hook load with 4-fall reeving. With four rope segments supporting the load block, each segment carries one quarter of the hook load, so the rope's SWL supports a hook load of

F_{\text{hook}} = 4 \times SWL = 4 \times 43 = 172 \text{ kN}

which is about 17.5 tonnes. The hook block, pin and any spreader must each be rated to at least 172 kN as well, or they become the weakest link.

Common Lifting Devices examiner traps

Counting rope segments on the fixed anchor block instead of the moving load block when finding mechanical advantage.
Adding gear ratios in series instead of multiplying them.
Forgetting efficiency: real gear stages and pulley systems waste energy as heat, so actual mechanical advantage is below the ideal value, and effort or motor torque is higher than the ideal calculation suggests.
Forgetting the distance trade-off: a force multiplied by a factor of $n$ requires the input to move $n$ times as far.
Using the breaking load directly as the working load, instead of dividing by the factor of safety; and forgetting that multiple falls share the hook load between rope segments.

Check your knowledge

Work through these under exam conditions and check against the solutions block. Show your working and carry units on every line.

Define mechanical advantage, velocity ratio and efficiency for a lifting machine, and state the relationship between them. (3 marks)
A single movable pulley is used to lift a 600 N load at an efficiency of 90 percent. Calculate (a) the velocity ratio, (b) the actual mechanical advantage, and (c) the effort force required. (4 marks)
A 2-and-2 block and tackle (4 supporting segments) lifts a 1000 N load at 80 percent efficiency. Calculate the effort force, and the length of rope pulled to raise the load 2.0 m. Verify your answer using a work-in equals work-out energy balance. (5 marks)
Two gear stages of 3:1 and 6:1 are connected in series. The input torque is 8.0 N m at 1500 rpm and the overall efficiency is 90 percent. Calculate (a) the overall gear ratio, (b) the output torque, and (c) the output speed in rpm. (5 marks)
A goods hoist must lift a 5000 N load at 0.50 m/s on a drum of radius 0.15 m. The overall gear efficiency is 75 percent and the motor runs at 1450 rpm. Calculate the required gear ratio and the motor torque required. (6 marks)
A hydraulic press has an input piston of diameter 25 mm and an output piston of diameter 150 mm. An input force of 300 N is applied. Calculate (a) the ratio of output force to input force, and (b) the output force. (4 marks)
A mobile crane uses wire rope with a minimum breaking load of 180 kN. AS1418 requires a factor of safety of 5. Calculate (a) the safe working load of a single rope and (b) the maximum hook load using a 3-fall reeving. (4 marks)
Explain why a worm-gear hoist is described as self-locking, and state why a separate mechanical brake is still required under AS1418. (4 marks)

Solutions

Q1: Mechanical advantage is the ratio of load force to effort force, $MA = F_{\text{load}}/F_{\text{effort}}$ . Velocity ratio is the ratio of effort distance (or speed) to load distance (or speed), $VR = d_{\text{effort}}/d_{\text{load}}$ , fixed by the machine geometry. Efficiency is the ratio of actual mechanical advantage to velocity ratio (equivalently output work over input work), $\eta = AMA/VR$ . The relationship is $AMA = \eta \times VR$ , so a real machine's mechanical advantage is always below its velocity ratio because of friction losses.
Q2: (a) A single movable pulley has two supporting segments, so $VR = 2$ . (b) $AMA = \eta \times VR = 0.90 \times 2 = 1.8$ . (c) $F_{\text{effort}} = F_{\text{load}}/AMA = 600/1.8 = 333$ N.
Q3: $IMA = VR = 4$ , $AMA = 0.80 \times 4 = 3.2$ . Effort $= 1000/3.2 = 312.5$ N. Rope pulled $= VR \times d_{\text{load}} = 4 \times 2.0 = 8.0$ m. Energy check: work in $= 312.5 \times 8.0 = 2500$ J, work out $= 1000 \times 2.0 = 2000$ J, efficiency $= 2000/2500 = 0.80$ . Consistent.
Q4: (a) $GR = 3 \times 6 = 18$ . (b) $T_{\text{out}} = T_{\text{in}} \times GR \times \eta = 8.0 \times 18 \times 0.90 = 129.6$ N m. (c) $\omega_{\text{out}} = \omega_{\text{in}}/GR = 1500/18 = 83.3$ rpm.
Q5: Drum torque $T_{\text{drum}} = F r = 5000 \times 0.15 = 750$ N m. Drum speed $\omega_{\text{drum}} = v/r = 0.50/0.15 = 3.33$ rad/s; in rpm $= 3.33 \times 60/(2\pi) = 31.8$ rpm. Required gear ratio $GR = \omega_{\text{motor}}/\omega_{\text{drum}} = 1450/31.8 = 45.6$ . Motor torque $T_{\text{motor}} = T_{\text{drum}}/(GR \times \eta) = 750/(45.6 \times 0.75) = 21.9$ N m, about 22 N m. (Check: motor power $= T_{\text{motor}}\,\omega_{\text{motor}} = 22 \times (1450 \times 2\pi/60) = 3.34$ kW, and $Fv/\eta = 5000 \times 0.50 / 0.75 = 3.33$ kW. Consistent.)
Q6: (a) Area ratio equals diameter ratio squared: $(150/25)^2 = 6^2 = 36$ . (b) $F_{\text{out}} = F_{\text{in}} \times 36 = 300 \times 36 = 10\,800$ N.
Q7: (a) $SWL = MBL/FoS = 180/5 = 36$ kN per rope. (b) With 3-fall reeving each of three segments carries one third of the hook load, so the maximum hook load $= 3 \times 36 = 108$ kN (about 11 tonnes), provided the hook block and fittings are also rated to 108 kN.
Q8: A worm-gear hoist is self-locking because the lead angle of the worm is smaller than the friction angle, so the worm can drive the gear wheel but the gear wheel cannot back-drive the worm. When power is removed the suspended load cannot turn the gear backward, so the load is held in place. A separate mechanical brake is still required under AS1418 because self-locking is a friction-dependent geometry property that can fail with wear, lubrication change or shock loading, and the standard does not accept it as the sole holding device for safety.