How to Calculate the Acceleration of a Moving Elevator

An elevator can be moving upward or downward while accelerating, slowing down, or traveling at constant velocity. Its direction of motion alone does not tell you its acceleration. To calculate the acceleration, you need either information about how its velocity changes or enough information to determine the net force on it.

The method below covers both approaches. Since no mass, force, velocity, or time values are specified, there is no single numerical acceleration to calculate.

Use the change in velocity

Average acceleration is the change in velocity divided by the time interval:

$$a_{\text{avg}}=\frac{v_2-v_1}{\Delta t}$$

Here, $v_1$ and $v_2$ are velocities, which include direction, and $\Delta t$ is the elapsed time. In SI units, velocity is measured in metres per second (m/s), time in seconds (s), and acceleration in metres per second squared (m/s²).

Pick a positive direction before entering the velocities. For example, if upward is positive, an upward velocity is positive and a downward velocity is negative. A negative acceleration then means acceleration downward; it does not necessarily mean the elevator is moving downward.

This calculation gives average acceleration over the interval. It equals the instantaneous acceleration at every moment only if acceleration is constant (or if the acceleration happens to be the same throughout).

Use Newton’s second law

If you know the forces acting on the elevator, apply Newton’s second law:

$$\sum F=ma$$

For a simple elevator model, the main vertical forces are cable tension $T$ upward and gravitational force $mg$ downward. Taking upward as positive gives:

$$T-mg=ma$$

so the elevator’s acceleration is

$$a=\frac{T-mg}{m}$$

The mass $m$ is measured in kilograms, and $g$ is approximately $9.8\,\text{m/s}^2$ near Earth’s surface. This model treats the elevator and its load as one system and ignores effects such as friction and the mass of the cable.

If the elevator is traveling at constant velocity, its acceleration is zero and the net force is zero. In this simple model, that means the cable tension equals the elevator’s weight, $T=mg$. The elevator can still be moving: zero acceleration means its velocity is not changing, not that it is at rest.

Worked example: find acceleration from cable tension

Suppose an elevator and its load have a total mass of $800\,\text{kg}$, and the cable pulls upward with a tension of $9{,}000\,\text{N}$. Take upward as positive.

First calculate the weight:

$$mg=(800\,\text{kg})(9.8\,\text{m/s}^2)=7{,}840\,\text{N}$$

The net upward force is $9{,}000-7{,}840=1{,}160\,\text{N}$. Therefore,

$$\begin{aligned} a&=\frac{T-mg}{m}\\ &=\frac{9{,}000\,\text{N}-7{,}840\,\text{N}}{800\,\text{kg}}\\ &=1.45\,\text{m/s}^2 \end{aligned}$$

The positive result means the acceleration is upward. The elevator could be moving upward and speeding up, or moving downward and slowing down; the force calculation determines acceleration, not velocity.

Worked example: find acceleration from motion data

Suppose the elevator’s velocity changes from $2.0\,\text{m/s}$ upward to $5.0\,\text{m/s}$ upward over $3.0\,\text{s}$. With upward positive, both velocities are positive:

$$a_{\text{avg}}=\frac{5.0-2.0}{3.0}=1.0\,\text{m/s}^2$$

The average acceleration is $1.0\,\text{m/s}^2$ upward. If instead the velocity changed from $2.0\,\text{m/s}$ upward to zero in $2.0\,\text{s}$, the average acceleration would be $(0-2.0)/2.0=-1.0\,\text{m/s}^2$: downward, even though the elevator was moving upward while it slowed.

What if velocity and time are unavailable?

If acceleration is constant and you know the initial and final velocities but not the time, displacement can be used instead:

$$v_2^2=v_1^2+2a\Delta y$$

Here $\Delta y$ is signed vertical displacement using the same positive direction. Rearranging gives $a=(v_2^2-v_1^2)/(2\Delta y)$, provided $\Delta y$ is nonzero. If you know only the elevator’s position at one instant, or only that it is moving, there is not enough information to determine its acceleration.

Common mistakes

• Assuming motion means acceleration: An elevator moving at steady speed has zero acceleration.
• Confusing direction of motion with direction of acceleration: An upward-moving elevator can accelerate downward if it is slowing down.
• Using speed instead of signed velocity: Speed has no direction. For a velocity-change calculation, represent direction with positive and negative signs.
• Using the weight as the net force: Weight is only one force. Include cable tension or other relevant forces before applying $F_{\text{net}}=ma$.
• Mixing sign conventions: If upward is positive, use that choice for every force, velocity, and displacement in the calculation.