Lift Motor: Cable Tension, Apparent Weight and When the Power Demand Peaks
A 1200 kg lift car carries four passengers of total mass 300 kg and is pulled by a single vertical cable. Starting from rest at the ground floor, it accelerates upward at 1.5 m/s² for 2.0 s, then travels at constant speed for 8.0 s, then decelerates uniformly and stops after a further 2.0 s at the tenth floor. Inside, a 60 kg student stands on a bathroom scale calibrated in newtons and watches the reading change. The motor is rated at 15 kW. Friction, air resistance and the mass of the cable are negligible. Take g = 10 m/s².
The building engineer says the motor must supply its greatest power at the instant the lift starts to move, because that is when the cable force is largest. A technician disagrees and says the greatest demand comes later in the journey. The student also reports that the scale reading near the top of the ride is lower than her normal weight.
1. Calculate the cable tension in each of the three phases and state the maximum speed reached.
Read the worked answer
Total mass (m = 1200 + 300 = 1500\text{ kg}), so weight (W = mg = 15000\text{ N}).
Phase 1 (upward acceleration 1.5 m/s²): the resultant force is upward, so (T - mg = ma). (T_1 = m(g+a) = 1500(10+1.5) = 17,250\text{ N})
Phase 2 (constant speed): acceleration is zero, so forces balance. (T_2 = mg = 15,000\text{ N})
Phase 3 (deceleration): it slows from the maximum speed to rest in 2.0 s. The maximum speed is (v = at = 1.5\times2.0 = 3.0\text{ m/s}), so the deceleration is (3.0/2.0 = 1.5\text{ m/s}^2) downward: (T_3 = m(g-a) = 1500(10-1.5) = 12,750\text{ N})
Check: the tension is greatest while accelerating and least while decelerating, which is exactly why the ride feels heavy at the start and light at the top. No calculation of tension should ever use the passenger's mass alone - the cable carries car plus passengers.
2. Predict the bathroom-scale reading during each phase and explain the student's observation near the top floor.
Read the worked answer
The scale reads the normal contact force N on the student (60 kg), not her weight.
Phase 1: (N = m(g+a) = 60\times11.5 = 690\text{ N}) (apparent weight ≈ 69 kg) - she feels heavier.
Phase 2: (N = mg = 600\text{ N}) - the scale reads her true weight, and she cannot tell she is moving at constant speed.
Phase 3: (N = m(g-a) = 60\times8.5 = 510\text{ N}) (≈ 51 kg) - she feels lighter, matching her report near the top.
Reasoning to state: the scale measures the force the floor pushes up on her. Her weight stays 600 N throughout; what changes is the acceleration, so the resultant force on her changes. During deceleration the resultant must point downward, so the upward normal force must be smaller than her weight. If the cable snapped and the car fell freely, (a = g) and N would fall to zero - the scale would read zero even though gravity still acts on her.
3. Decide who is right about when the motor's power demand peaks, supporting your answer with calculations.
Read the worked answer
Both are partly wrong in their reasoning, and the numbers settle it.
Mechanical power delivered to the lift is (P = Tv), where v is the instantaneous speed.
At the instant of starting: the tension is at its maximum (17 250 N) but (v = 0), so (P = 17,250\times0 = 0). The engineer's reasoning is therefore wrong - a large force with no movement does no work.
End of the acceleration phase ((v = 3.0\text{ m/s})): (P = 17,250\times3.0 = 51,750\text{ W} \approx 52\text{ kW}).
Constant-speed phase: (P = 15,000\times3.0 = 45,000\text{ W} = 45\text{ kW}).
During deceleration: power output falls as v falls and T is smallest, so it is lower still.
So the peak demand occurs at the end of the acceleration phase (≈52 kW), where a large tension is multiplied by the largest speed - the technician is right that it is not at the start, but the peak is during acceleration, not during the constant-speed cruise.
Consequence: the 15 kW rating is far below the 52 kW peak, so it cannot be the mechanical output rating alone; it must be interpreted as a continuous/average rating, or the motor must be geared with a counterweight so the cable force is much smaller than the full load weight. This is exactly why real lifts use counterweights.