Hoist Gear Train, Drum Speed and Pawl Safety
A workshop hoist is driven by an electric motor through a two-stage gear train. The motor shaft carries a 12-tooth gear, which meshes with a 48-tooth gear. The 48-tooth gear is on the same shaft as a 15-tooth gear, which meshes with a 45-tooth gear. The 45-tooth gear is fixed to a rope drum of diameter 200 mm. A single rope winds onto the drum and lifts a 300 kg load vertically. The motor runs at 1,440 rpm and delivers 1.5 kW at the motor shaft. Assume the gears and rope are lossless for the first calculation. A spring-applied pawl engages a ratchet on the drum shaft; the pawl is released by a solenoid only while the 'raise' button is held. An operator suggests removing the pawl because the gear train is 'not back-drivable anyway'.
1. Calculate the drum speed in rpm and the load's lifting speed in m/s.
Read the worked answer
Overall ratio = (48 ÷ 12) × (45 ÷ 15) = 4 × 3 = 12. Drum speed = 1,440 ÷ 12 = 120 rpm. Drum circumference = π × 0.200 = 0.6283 m. Rope speed = 120 × 0.6283 = 75.40 m/min = 1.257 m/s.
The load rises at about 1.26 m/s when the motor runs at full speed. The two-stage arrangement multiplies speed reduction: the first pair alone would give 360 rpm, and the second pair takes that down by a further factor of three.
2. Calculate the motor torque required to raise the 300 kg load at constant speed, and compare it with the torque available from 1.5 kW at 1,440 rpm. What do you conclude?
Read the worked answer
Load force = 300 × 9.81 = 2,943 N. Drum torque = force × drum radius = 2,943 × 0.100 = 294.3 N·m. Motor torque required = 294.3 ÷ 12 = 24.5 N·m. Torque available from 1.5 kW at 1,440 rpm: angular speed = 1,440 × 2π ÷ 60 = 150.8 rad/s; torque = 1,500 ÷ 150.8 = 9.95 N·m.
The motor provides only about 40% of the torque needed. At 1.257 m/s the maximum load is 9.95 × 12 ÷ 0.100 ÷ 9.81 ≈ 122 kg. The 300 kg rating and the 1.5 kW motor are inconsistent; check whether the drum is smaller, the speed is lower, or the motor is larger than stated.
3. Assess the operator's proposal to remove the pawl. What should you do?
Read the worked answer
Do not remove the pawl. The 'not back-drivable' claim is not supported: a lossless spur-gear train is reversible, and the load's torque at the drum (294 N·m for 300 kg) is far greater than the torque the motor can hold at the drum (9.95 × 12 = 119 N·m). If the motor loses power or the drive slips, the load can run down. A pawl is a positive mechanical hold that does not depend on power, friction or a brake.
Keep the pawl and its solenoid interlock, verify the hoist's rated load against the actual 300 kg, and ask the supplier to confirm the drum diameter, gear ratio and motor rating. If the pawl is nuisance-tripping, fix the solenoid or ratchet rather than removing the safety device.