Elevator Power Calculation
This simulation demonstrates the power required by an elevator moving upward at constant speed against gravity and friction, based on Example 6.11 from the textbook.
Total Force
22000 N
Power Required
44000 W (59 hp)
Height
0 m
Time
0 s
Motor Force (Fmotor)
Gravity (mg)
Friction (Ff)
Physics Explanation:
From the textbook example with a 1800 kg elevator moving at 2 m/s against 4000 N friction:
Total downward force:
F = mg + Ff = (1800 kg × 10 m/s²) + 4000 N = 22000 N
F = mg + Ff = (1800 kg × 10 m/s²) + 4000 N = 22000 N
Power required to maintain constant speed:
P = F × v = 22000 N × 2 m/s = 44000 W
In horsepower: 44000 W ÷ 746 W/hp ≈ 59 hp
P = F × v = 22000 N × 2 m/s = 44000 W
In horsepower: 44000 W ÷ 746 W/hp ≈ 59 hp
The motor must supply exactly enough power to balance the combined forces of gravity and friction to maintain constant velocity.