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Elevators - Force and Power

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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 (h 1 - h 0 )                     (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)

h 1 = final elevation (m, ft)

h 0 = 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 (h 1 - h 0 )              (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 (v 1 - v 0 ) / t a (5)

where

F a = acceleration force (N, lb f )

v 1 = final velocity (m/s, ft/s)

v 0 = 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 (h 1 - h 0 ) / 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

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