The Physics of a Falling Lift
A lift of mass 800 kg (including passengers) is suspended by a steel cable. During a test, the cable is cut and the lift falls freely from rest for 1.5 s before an emergency brake engages and brings it to rest in a further 0.4 s. A safety inspector asks whether the brake's deceleration would injure a standing passenger. The inspector notes that the lift has a maximum rated deceleration of 1.5 g, where g is about 9.8 m/s². The brake manufacturer claims the brake can stop the lift in 0.3 s, but the test showed 0.4 s. The inspector also asks what would happen if the cable had not been fully cut and the lift had fallen more slowly. Assume air resistance is negligible.
1. Calculate the speed of the lift when the brake engages, and the deceleration during braking. Compare the deceleration with the rated limit.
Read the worked answer
Speed at brake engagement. Free fall from rest for 1.5 s: v = g t = 9.8 × 1.5 = 14.7 m/s (about 53 km/h).
Deceleration during braking. The lift goes from 14.7 m/s to rest in 0.4 s: a = Δv / t = 14.7 / 0.4 = 36.75 m/s².
As a multiple of g: 36.75 / 9.8 ≈ 3.75 g.
Comparison. The rated maximum is 1.5 g, so the measured deceleration is about 2.5 times the limit. The brake as tested would subject a standing passenger to roughly 3.75 times their body weight, which is well beyond the rated safe level.
Note on the manufacturer's claim. At 0.3 s the deceleration would be 14.7 / 0.3 = 49 m/s² ≈ 5 g, which is even worse. A shorter stopping time increases deceleration, so the manufacturer's claim is not reassuring - it points the wrong way. The brake is not compliant as tested.
2. The inspector asks what would happen if the cable had not been fully cut and the lift had fallen more slowly. Explain how the braking deceleration would change and why.
Read the worked answer
If the lift fell more slowly, it would reach the brake with a lower speed, and the deceleration during braking would be lower for the same braking time.
- Speed scales with fall. If the fall is slower - say the lift only reached 7 m/s instead of 14.7 m/s - then over the same 0.4 s braking time the deceleration would be 7 / 0.4 = 17.5 m/s² ≈ 1.8 g, closer to the 1.5 g limit.
- The relationship is linear. Deceleration = speed at engagement ÷ braking time, so halving the speed halves the deceleration, provided the brake acts over the same time.
- But the brake's behaviour may change. A brake that produces a roughly constant force will stop a slower lift in a shorter time, not the same time. In that case the deceleration could be similar or the stopping distance shorter. The outcome depends on whether the brake is time-limited or force-limited.
- The safety implication. The worst case is the fastest fall, which is the fully cut cable. Designing for the slower fall would understate the hazard.
The inspector should require the brake to be rated for the worst-case engagement speed, not the typical one.
3. Using the lift's mass, estimate the force the cable must have supported before it was cut, and the force the brake must exert during the test. State any assumptions.
Read the worked answer
Before the cut (static). The cable supports the weight: F = m g = 800 × 9.8 = 7,840 N ≈ 7.8 kN.
Assumption: the lift is at rest or moving at constant speed, so the only vertical force is weight.
During braking. The brake must provide the net upward force to decelerate the lift at 36.75 m/s²: F_net = m a = 800 × 36.75 = 29,400 N ≈ 29.4 kN.
This net force is the brake force minus the weight (which still acts downward). So the brake force is: F_brake = F_net + m g = 29,400 + 7,840 = 37,240 N ≈ 37 kN.
Assumptions and caveats.
- The mass of 800 kg is taken as constant; the cable's own mass is neglected.
- Air resistance is neglected, as stated.
- The deceleration is assumed uniform, which gives an average force; the peak force may be higher.
- The brake force is about 4.75 times the static cable load, which is why the brake and its mountings must be rated for far more than the lift's weight.
This estimate shows the structural demand on the brake is much greater than the everyday load.