HSC Engineering Studies Lifting Devices: deep-dive 2026 guide

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:

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

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.

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):

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:

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 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):

Output force:

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.

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:

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

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.

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.