# Elevators - Force and Power

### 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/s^{2}, 32.17 ft/s^{2}) *

* 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/s^{2}) *

* = 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/s ^{2}) ((15 m) - (0 m)) / (20 s) *

* = 14865 W *

* = 14.9 kW *

## Related Topics

### • Dynamics

Motion - velocity and acceleration, forces and torque.

### • Mechanics

Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.

## Related Documents

### Acceleration of Gravity and Newton's Second Law

Acceleration of gravity and Newton's Second Law - SI and Imperial units.

### Factors of Safety - FOS

Factors of Safety - FOS - are important in engineering designs.

### Formulas of Motion - Linear and Circular

Linear and angular (rotation) acceleration, velocity, speed and distance.

### Hot Air Balloons - Calculate Lifting Weights

Calculate hot air ballons lifting forces.

### Levers

Use levers to magnify forces.

### Power

Power is the rate at which work is done or energy converted.

### Pulleys

Pulleys, blocks and tackles.

### Torque - Work done and Power Transmitted

The work done and power transmitted by a constant torque.

### Wire Ropes - Strengths

6 strand x 19 wire (6x19) - minimum breaking strength, safe loads and weight.