Elevator Force Power Lift

Overview

Engineering reference data for Elevator Force Power Lift in mechanics.

Key Formulas

Newton's Second Law

$F = ma$

Force = mass × acceleration.

Work

$W = Fd\cos\theta$

Work = force × displacement × cos(angle).

Kinetic Energy

$E_k = \frac{1}{2}mv^2$

Energy of motion.

Potential Energy

$E_p = mgh$

Gravitational potential energy.

Variables

Symbol	Description	Unit
$F$	Force	N
$m$	Mass	kg
$a$	Acceleration	m/s²
$v$	Velocity	m/s

Force During Acceleration and Deceleration

When an elevator starts or stops, additional forces act on the system beyond the constant speed force. The acceleration force can be expressed as:

$F_a = m \cdot \frac{v_1 - v_0}{t_a}$

where:

• $F_a$ = acceleration force (N, lbf)
• $v_1$ = final velocity (m/s, ft/s)
• $v_0$ = initial velocity (m/s, ft/s)
• $t_a$ = acceleration or deceleration time (s)

During startup ($v_0 = 0$), the total force required is:

$F_{total} = F_c + F_a = m \cdot g + m \cdot \frac{v_1}{t_a}$

During stopping ($v_1 = 0$), the braking force reduces the required cable tension:

$F_{braking} = F_c - F_a = m \cdot g - m \cdot \frac{v_0}{t_a}$

Safety Note: If deceleration exceeds gravitational acceleration ($a > g$), passengers experience upward forces and may feel momentarily weightless or experience discomfort.

Worked Example

An elevator system with the following parameters:

Parameter	Value
Mass (including passengers)	2000 kg
Initial elevation	0 m
Final elevation	15 m
Travel time	20 s

Constant speed force:

$F_c = 2000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19{,}620 \, \text{N} \approx 19.6 \, \text{kN}$

Power required:

$P = \frac{2000 \times 9.81 \times (15 - 0)}{20} = 14{,}715 \, \text{W} \approx 14.7 \, \text{kW}$

Average velocity:

$v_{avg} = \frac{15 \, \text{m}}{20 \, \text{s}} = 0.75 \, \text{m/s}$

References

• Original Source