The Drawbridge That Stalls and Falls
A student builds a motorized model drawbridge for the science fair. A 6.0 V battery pack (four AA cells) drives a DC motor whose shaft carries a 12-tooth pinion meshing with a 60-tooth gear fixed to a winch drum 4.0 cm in diameter. A string wound on the drum lifts the 400 g bridge flap. The motor's data sheet gives 0.020 N·m of torque at 6.0 V and a stall current of 1.2 A. The bridge rises too fast, so the student solders a 10 Ω resistor in series with the motor; the ammeter now reads 0.20 A. After ten lifts the resistor is too hot to touch, the flap rises more slowly than before, and the motor sometimes stalls. When the motor is switched off, the flap falls back down. The teacher says the resistor is "the wrong tool for the job" and that the mechanism needs a self-locking feature.
1. Using the readings, find the voltage across the resistor and across the motor, the power wasted, and decide whether the motor can still lift the 400 g flap.
Read the worked answer
Voltage across the resistor: V = IR = 0.20 A × 10 Ω = 2.0 V. Voltage left for the motor: 6.0 V − 2.0 V = 4.0 V.
Power: the battery delivers P = VI = 6.0 V × 0.20 A = 1.2 W. The resistor dissipates P = I²R = (0.20)² × 10 = 0.40 W, and the motor receives 4.0 V × 0.20 A = 0.80 W. So one third of the energy is being converted to heat in the resistor - that is why it is hot.
Torque check: with only 4.0 V applied, the motor's torque falls roughly in proportion, to about 0.020 × (4.0/6.0) = 0.013 N·m. The 12:60 gear pair multiplies torque by 60/12 = 5, giving about 0.067 N·m at the drum. The drum radius is 2.0 cm = 0.020 m, so the available lifting force is 0.067 ÷ 0.020 ≈ 3.3 N.
The load needs F = mg = 0.400 kg × 9.8 N/kg ≈ 3.9 N. Since 3.3 N < 3.9 N, the motor cannot lift the flap - it stalls, which is exactly what the student observes.
2. What is the gear train doing, and how should the bridge be slowed without wasting energy?
Read the worked answer
The 12-tooth pinion driving the 60-tooth gear is a 5:1 speed reduction. The drum turns once for every five turns of the motor, so the lifting speed drops by a factor of 5 while the torque at the drum rises by the same factor (minus friction losses). That trade of speed for torque is what lets a small motor lift a load at all.
To slow the bridge further, change the mechanism instead of burning energy:
- fit a larger driven gear, e.g. 90 teeth for 7.5:1, or use a smaller drum - both reduce lifting speed and keep the torque margin;
- if the supply must change, use fewer cells or a pulse-width-modulated speed controller, which delivers only the energy the motor actually uses.
A series resistor is a poor solution for two reasons: it converts a fixed fraction of the energy into heat, and its voltage drop changes with the current, so the torque available when the load is heaviest is exactly when it is worst. The test of any fix is whether the loaded flap still lifts with margin.
3. Explain why the flap drops when the power is cut, and propose a self-locking solution.
Read the worked answer
The gear train is reversible. With the motor off, the load on the drum can drive the 60-tooth gear backwards, turning the 12-tooth pinion and the motor shaft. Nothing resists that motion except friction, so the 400 g flap descends.
Possible fixes, in order of suitability for this model:
- Worm gear: a rotating worm driving a wheel cannot normally be back-driven because of the shallow lead angle. It is self-locking and also gives a large reduction in one stage.
- Ratchet and pawl: a toothed wheel with a spring-loaded pawl allows lifting in one direction only. Simple to build, but noisy and it must be released to lower the bridge.
- Friction brake or counterweight: a light brake on the drum, or a counterweight that balances the flap so it stays where it is put.
Any solution should be tested the same way: switch off with the 400 g load at mid-height and check that the flap holds its position. A mechanical stop should also be fitted so the flap cannot overshoot.