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

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:

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