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2.2.6.4. Elevators

Table of Contents

Motion and Forces in Elevators

An elevator is a very useful example of Newton's laws because it shows how the same person can feel heavier, lighter, or even weightless, depending on the elevator's acceleration. The actual gravitational force on the person does not usually change during the ride. What changes is the support force from the floor, which is the normal force.

When a person stands on the floor of an elevator, two main vertical forces act on the person. Gravity pulls downward with force $mg$, and the floor pushes upward with normal force $N$. If the elevator accelerates, these forces do not balance in the same way as when the elevator is at rest.

Choosing a Sign Convention

To apply Newton's second law clearly, choose a positive direction. A common choice is upward as positive. Then the vertical force equation for a person in the elevator is

$$
\sum F_y = ma
$$

which becomes

$$
N - mg = ma
$$

Here, $a$ is the acceleration of the person, which is the same as the acceleration of the elevator, as long as the person remains in contact with the floor.

Solving for the normal force gives

$$
N = m(g + a)
$$

This one equation explains the main situations in elevator problems.

Important elevator equation, with upward chosen as positive:
$$
N - mg = ma
$$
So,
$$
N = m(g + a)
$$
The apparent weight of a person in the elevator is the normal force $N$, not the true weight $mg$.

Apparent Weight

A scale in an elevator does not measure the true gravitational force directly. It measures the force the floor exerts on the person. This is the normal force, and it is often called the apparent weight.

If the elevator accelerates upward, then $a > 0$, so

$$
N > mg
$$

The person feels heavier.

If the elevator accelerates downward, then $a < 0$, so

$$
N < mg
$$

The person feels lighter.

If the elevator has no acceleration, then $a = 0$, so

$$
N = mg
$$

This happens when the elevator is at rest or moving with constant velocity.

Common Elevator Cases

The main cases can be summarized clearly.

Elevator motionAcceleration $a$Normal force $N$Feeling
At rest$0$$mg$Normal
Moving upward at constant speed$0$$mg$Normal
Moving downward at constant speed$0$$mg$Normal
Accelerating upwardpositivegreater than $mg$Heavier
Accelerating downwardnegativeless than $mg$Lighter

Notice that velocity alone does not change apparent weight. Only acceleration changes it.

A person does not feel heavier because the elevator is moving upward.
A person feels heavier only if the elevator is accelerating upward.
Similarly, moving downward does not make a person feel lighter unless there is downward acceleration.

Weightlessness in an Elevator

A special case happens when the elevator accelerates downward with magnitude $g$. Then

$$
a = -g
$$

and the force equation gives

$$
N = m(g - g) = 0
$$

If the normal force becomes zero, the floor no longer pushes on the person. This is apparent weightlessness.

This situation is like free fall. The person is still under gravity, but there is no support force.

Apparent weightlessness means
$$
N = 0
$$
In an elevator, this occurs when the elevator and person accelerate downward at $g$.

Tension in the Elevator Cable

If we study the whole elevator car instead of the person inside it, the same method applies. Suppose the elevator of mass $M$ is pulled upward by cable tension $T$, while gravity pulls downward with force $Mg$. If upward is positive, then

$$
T - Mg = Ma
$$

So,

$$
T = M(g + a)
$$

This looks just like the equation for the normal force on a person. If the elevator accelerates upward, the cable tension must be greater than the elevator's weight. If it accelerates downward, the tension is less than the weight.

In some problems, the total mass includes the elevator plus passengers. Then use the combined mass in the equation.

A Person Standing on a Scale

A very common elevator problem involves a person standing on a scale. The scale reading equals the normal force. If the person's mass is $m$, then

$$
\text{scale reading} = N
$$

If the elevator accelerates upward at $a$,

$$
N = m(g + a)
$$

If the elevator accelerates downward with magnitude $a$,

$$
N = m(g - a)
$$

where here $a$ is taken as a positive number representing the size of the downward acceleration.

This form is often easier for beginners because it separates upward and downward cases.

Example Calculation

Suppose a person of mass $70 \, \text{kg}$ stands in an elevator accelerating upward at $2.0 \, \text{m/s}^2$. Take $g = 9.8 \, \text{m/s}^2$.

Then

$$
N = m(g + a)
$$

so

$$
N = 70(9.8 + 2.0) = 70(11.8) = 826 \, \text{N}
$$

The person's true weight is

$$
mg = 70 \times 9.8 = 686 \, \text{N}
$$

So the scale reads $826 \, \text{N}$, which is greater than the true weight. The person feels heavier.

Now suppose instead the elevator accelerates downward at $2.0 \, \text{m/s}^2$. Then

$$
N = m(g - a)
$$

so

$$
N = 70(9.8 - 2.0) = 70(7.8) = 546 \, \text{N}
$$

Now the scale reads less than the true weight, so the person feels lighter.

Force Diagram

A simple force diagram helps organize elevator problems.

Person standing in an accelerating elevator

The person has an upward normal force and a downward weight. The direction of the acceleration depends on which force is larger.

Starting, Stopping, and Constant Speed

During a real elevator trip, the acceleration often changes in stages. At the beginning, the elevator may accelerate upward. Then the rider feels heavier. After that, the elevator may move upward at constant speed. Then the rider feels normal again because $a = 0$. Finally, as the elevator approaches the destination above, it must slow down. If it is moving upward but slowing down, the acceleration is downward, so the rider feels lighter.

This is an important idea. The direction of acceleration is not always the same as the direction of motion.

For a downward trip, the same logic applies. At first, the elevator may accelerate downward, making the rider feel lighter. Later it may move downward at constant speed, giving a normal feeling. Near the bottom, it slows down, so the acceleration is upward, and the rider feels heavier.

Interpreting the Signs Carefully

Many mistakes in elevator problems come from sign confusion. The safest method is to define a positive direction first, write the forces with signs, and then use Newton's second law exactly.

For upward positive,

$$
N - mg = ma
$$

If $a$ comes out positive, the acceleration is upward. If $a$ comes out negative, the acceleration is downward.

If you instead choose downward as positive, the equation changes form, but the physics stays the same. What matters is consistency.

In elevator problems, do not guess from motion alone.
Always base the analysis on acceleration and Newton's second law.

Elevator as a System

Sometimes it is useful to consider the person and elevator together as one system. Internal forces between the person and the floor then cancel inside the system. External forces, such as cable tension and gravity, determine the acceleration. This can simplify more advanced problems, especially when the mass of the elevator car and the passenger are both included.

Practical Meaning

Elevator problems are important because they connect equations to real sensations. The force you feel under your feet is not always equal to your true weight. Your body senses the support force, and that is why acceleration changes how heavy or light you seem.

This makes elevators one of the clearest everyday examples of Newton's second law in action.

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