Scale reading and a helium balloon in a descending lift
A 60 kg student stands on a spring scale calibrated in newtons inside a hospital lift. The lift starts from the 8th floor and descends to the 2nd floor. At the instant it begins to move the scale reads 540 N and holds that value for 4 s. The reading then changes to 660 N and stays there for 4 s before the lift comes to rest. Take g = 10 m/s² and ignore the mass of the scale. A helium-filled balloon is tied by a light string to a ring on the floor of the lift. A co-passenger insists that the 540 N reading proves the lift is moving downward and that 660 N proves it must be moving upward. The student has to write a short note for her practical file explaining what the scale really measures, what the balloon's string does during each phase, and what would happen to the scale reading and the balloon if the cable snapped while the lift was descending at 3 m/s.
1. Calculate the acceleration of the lift in each of the two phases. Given that the lift is known to be descending, decide whether it is speeding up or slowing down in each phase, and identify exactly where the co-passenger's reasoning fails.
Read the worked answer
The scale reads the normal force N. Taking upward as positive, N − mg = ma, so a = N/m − g.
- Phase 1: a = 540/60 − 10 = 9 − 10 = −1 m/s², i.e. 1 m/s² downward.
- Phase 2: a = 660/60 − 10 = 11 − 10 = +1 m/s², i.e. 1 m/s² upward.
The velocity is downward throughout, because the lift goes from the 8th to the 2nd floor. Downward velocity with downward acceleration means the speed is increasing; downward velocity with upward acceleration means the speed is decreasing. The times are self-consistent: speed rises from 0 to 4 m/s in 4 s, then falls back to 0 in the next 4 s.
The co-passenger has confused acceleration with velocity. A spring balance measures N and therefore only a; it says nothing about the direction of v. The 660 N reading would also be obtained in a lift that is ascending and speeding up. Only the floor numbers, not the scale, fix the direction of motion.
2. Explain what happens to the tension in the balloon's string in each phase, and state what the student would observe if the cable snapped.
Read the worked answer
In the frame of the cabin the air behaves like a fluid with effective gravity (g + a), so the upthrust is ρ_air V (g + a) and the balloon's weight is ρ_He V (g + a). The string tension is
T = (ρ_air − ρ_He) V (g + a), with a positive upward.
- Phase 1 (a = −1 m/s²): g + a = 9 m/s², so T falls to about 90% of its value when the lift is at rest; the balloon hangs slightly lower and the string is less taut.
- Phase 2 (a = +1 m/s²): g + a = 11 m/s², so T rises by about 10% and the string tightens.
If the cable snapped, a = −g and effective gravity becomes zero. Buoyancy disappears because the cabin air is itself in free fall, the string goes slack, the scale reads zero, and the balloon does not rush upward - it stays beside the student, both falling together. The downward speed of 3 m/s at the moment of snapping is irrelevant to all of these readings.
3. What would the scale read, and what would the balloon do, if the lift later moved upward at a steady 3 m/s?
Read the worked answer
Steady velocity means a = 0, so N = mg = 600 N and the scale reads exactly the student's true weight. The balloon's string returns to its at-rest tension (ρ_air − ρ_He)Vg. This is the key contrast with the earlier phases: motion itself never changes a balance reading, only acceleration does. A candidate who answers 660 N here has carried over the acceleration from the previous phase instead of re-reading the new condition.