Elevators - Force and Power

Required force and power to lift an elevator.

Work done by Lifting the Elevator

The work done by lifting an elevator from one level to an other can be expressed as

W = m a_g (h₁ - h₀) (1)

where

W = work done (J, ft lb_f)

m = mass of elevator and passengers (kg, lb_m)

a_g = acceleration of gravity (9.81 m/s², 32.17 ft/s²)

h₁ = final elevation (m, ft)

h₀ = initial elevation (m, ft)

The generic equation for work done by a force can be expressed as

W = F_c s (2)

where

W = work done by force (J, ft lb_f)

F_c = force acting on the elevator at constant speed (N, lb_f)

s = distance moved by elevator (m, ft)

Forces acting on the Elevator

Since works done in (1) and (2) are equal - the equations can be combined to

F_c s = m a_g (h₁ - h₀) (3)

Force at constant Speed

Since the difference in elevation and the distance moved by the force are equal - (3) can be modified to express the force required to move the elevator at constant speed to

F_c = m a_g (4)

Force at start/stop

When the elevator starts or stops - the acceleration or deceleration force in addition to the constant speed force can be expressed as

F_a = m (v₁ - v₀) / t_a (5)

where

F_a = acceleration force (N, lb_f)

v₁ = final velocity (m/s, ft/s)

v₀ = initial velocity (m/s, ft/s)

t_a = start or stop (acceleration) time (s)

Power required to move the Elevator

The power required to move the elevator can be calculated as

P = W / t

= m a_g (h₁ - h₀) / t (6)

where

P = power (W, ft lb_f)

t = time to move the elevator between levels (s)

Example - Force and Power to Lift an Elevator

An elevator with mass 2000 kg including passengers are moved from level 0 m to level 15 m. The force required to move the elevator at constant speed can be calculated as

F_c = (2000 kg) (9.81 m/s²)

= 19820 N

The power required to move the elevator between the levels in 20 s can be calculated as

P = (2000 kg) (9.81 m/s²) ((15 m) - (0 m)) / (20 s)

= 14865 W

= 14.9 kW