Understanding Apparent Weight in an Elevator During Acceleration

When a person stands on a spring weighing machine inside an elevator, the reading can change depending on the motion of the elevator. Specifically, when the elevator ascends or descends, the apparent weight differs from the person's actual weight. In this article, we will explore how to calculate the apparent weight during the acceleration of the elevator.

Apparent Weight Formula

The apparent weight $F_{text{apparent}}$ of a person standing on a spring weighing machine in an accelerating elevator can be calculated using the formula:

$F_{text{apparent}} (m times g) (m times a)$

Where $m$ is the mass of the person, $g$ is the acceleration due to gravity, and $a$ is the acceleration of the elevator.

Example Calculation

Let's apply this formula to a scenario where a person with a mass of 80 kg stands on a spring weighing machine inside an elevator that starts to ascend with an acceleration of 2.0 m/s2:

Substituting these values into the formula, we get:

$F_{text{apparent}} (80 , text{kg} times 9.81 , text{m/s}^2) (80 , text{kg} times 2.0 , text{m/s}^2)$

$F_{text{apparent}} (80 , text{kg} times 11.81 , text{m/s}^2)$

$F_{text{apparent}} 944.8 , text{N}$

Therefore, the reading on the spring weighing machine when the elevator ascends with an acceleration of 2.0 m/s2 is approximately 944.8 N.

Special Cases

It's also worth noting that the apparent weight can change with variations in the acceleration due to gravity $g$. Let's consider some special cases:

When $g 10 , text{m/s}^2$

Using the same mass and acceleration of the elevator, the calculation would be:

$F_{text{apparent}} (80 , text{kg} times 10 , text{m/s}^2) (80 , text{kg} times 2.0 , text{m/s}^2)$

$F_{text{apparent}} 1000 , text{N}$

This means the reading on the weighing machine would be 1000 N when the acceleration due to gravity is 10 m/s2.

When $g 15 , text{m/s}^2$

Similarly, for $g 15 , text{m/s}^2$, the calculation is:

$F_{text{apparent}} (80 , text{kg} times 15 , text{m/s}^2) (80 , text{kg} times 2.0 , text{m/s}^2)$

$F_{text{apparent}} 1360 , text{N}$

The reading on the weighing machine would be 1360 N under these conditions.

Conclusion

In summary, the apparent weight of a person standing on a spring weighing machine is impacted by the motion of the elevator. Through the formula $F_{text{apparent}} (m times g) (m times a)$, we can accurately calculate the weight reading during different scenarios. This knowledge can be useful in understanding the mechanics involved in elevators and other similar systems.