The weight, normal force, and apparent weight calculator is based on Newton’s second law applied to a person, scale, or object inside an accelerating frame such as an elevator. The important idea is that true weight and apparent weight are not always the same. True weight is the gravitational force \(mg\), while apparent weight is the support force that a scale or floor exerts on the object.
1. True weight
The true weight of an object near a planet’s surface is the gravitational force:
\[
W = mg
\]
Here, \(m\) is the mass and \(g\) is the local gravitational acceleration. On Earth, a common value is:
\[
g \approx 9.81\,\mathrm{m/s^2}
\]
For a \(70\,\mathrm{kg}\) person, the true weight is:
\[
W = (70)(9.81) = 686.7\,\mathrm{N}
\]
2. Normal force and apparent weight
A scale does not directly measure gravitational force. It measures the normal force pressing upward on the person. This normal force is what we call apparent weight.
If upward is chosen as the positive direction, the forces on a person in an elevator are:
- normal force \(N\) upward,
- weight \(mg\) downward.
Newton’s second law gives:
\[
\sum F_y = ma
\]
\[
N - mg = ma
\]
Solving for \(N\):
\[
N = m(g+a)
\]
This is the key formula for elevator apparent-weight problems.
3. Elevator accelerating upward
If the elevator accelerates upward, then \(a>0\). The normal force becomes larger than true weight:
\[
N = m(g+a) > mg
\]
For example, with \(m=70\,\mathrm{kg}\), \(g=9.81\,\mathrm{m/s^2}\), and \(a=2\,\mathrm{m/s^2}\):
\[
\begin{aligned}
N &= 70(9.81+2) \\
&= 70(11.81) \\
&= 826.7\,\mathrm{N}
\end{aligned}
\]
The person feels heavier because the floor must push upward with more force than usual.
4. Elevator accelerating downward
If the elevator accelerates downward, then \(a<0\). The normal force becomes smaller:
\[
N = m(g+a) < mg
\]
For example, if \(a=-2\,\mathrm{m/s^2}\):
\[
\begin{aligned}
N &= 70(9.81-2) \\
&= 70(7.81) \\
&= 546.7\,\mathrm{N}
\end{aligned}
\]
The person feels lighter because the floor does not need to push as hard to produce the downward acceleration.
5. Constant velocity and rest
If the elevator is at rest or moving at constant velocity, then acceleration is zero:
\[
a = 0
\]
The formula becomes:
\[
N = m(g+0) = mg
\]
Therefore, the apparent weight equals the true weight. This is true even if the elevator is moving upward or downward at constant speed. Velocity alone does not change the scale reading; acceleration does.
6. Free fall
In free fall, the elevator and the person accelerate downward at:
\[
a = -g
\]
Then:
\[
N = m(g-g)=0
\]
This produces apparent weightlessness. The person still has true weight \(mg\), but there is no support force from the floor. A scale would read zero.
7. Negative normal force and loss of contact
The raw formula \(N=m(g+a)\) can become negative if \(a<-g\). In a real contact situation, the floor or scale cannot pull downward on the person. It can only push upward. Therefore, a negative raw normal force means the person loses contact with the floor.
In that case, the physical normal force is:
\[
N_{\text{physical}} = 0
\]
The calculator can show either the raw formula value or the physical contact-limited value, depending on the selected output convention.
8. Apparent g-force
Apparent g-force compares the normal force to the true weight:
\[
\text{g-force} = \frac{N}{mg}
\]
If this value is \(1\), the person feels normal. If it is greater than \(1\), the person feels heavier. If it is between \(0\) and \(1\), the person feels lighter. If it is \(0\), the person is weightless.
| Situation |
Acceleration |
Normal force |
Feeling |
| At rest |
\(0\) |
\(N=mg\) |
Normal |
| Constant velocity |
\(0\) |
\(N=mg\) |
Normal |
| Accelerating upward |
\(a>0\) |
\(N>mg\) |
Heavier |
| Accelerating downward |
\(-g
\(0
| Lighter |
| |
| Free fall |
\(a=-g\) |
\(N=0\) |
Weightless |
9. Common mistakes
-
Confusing weight and apparent weight:
true weight is gravitational force, while apparent weight is the normal force.
-
Using velocity instead of acceleration:
constant velocity does not change apparent weight.
-
Using the wrong sign:
upward acceleration is positive in the formula \(N=m(g+a)\).
-
Forgetting free fall:
in free fall, \(a=-g\), so \(N=0\), not \(mg\).
-
Allowing a physical negative normal force:
a floor can push but cannot pull, so loss of contact gives \(N=0\).
10. Summary
Elevator apparent-weight problems are direct applications of Newton’s second law. Draw the forces, choose upward as positive, write \(N-mg=ma\), and solve for \(N\). The normal force is the apparent weight. The true weight \(mg\) remains the same for a fixed mass and gravity, but the scale reading changes whenever the elevator accelerates.
