Have you ever been in an elevator and for a moment felt lighter or heavier than normal? That's because the elevator during those times is accelerating and making your apparent weight different that your actual weight. In each of the situations below, what would the apparent weight be in comparison to the actual weight? Click on each image to check your understanding.
Question
What do you notice about the force of gravity in each situation?
The magnitude of the force of gravity is the same in each scenario. The mass of the woman does not change, therefore the force of gravity on him will also not change. The normal force is the one to change magnitude.
If you know what the mass of the woman that is in the elevator, you can figure out the magnitude of the normal force, also known as the apparent weight. Since the net force is equal to mass times acceleration and the net force is the sum of the forces in the y-direction, you can find the normal force. If the woman is 80.0 kg, find the apparent weight in each scenario. Click on the answer button below to check your work.
| Constant Speed | Upward Acceleration |
|---|---|
| \(\small\mathsf{ m = 80.0 \text{ kg} }\) \(\small\mathsf{ \overrightarrow{a} = 0.0 \text{ m/s}^2 }\) \(\small\mathsf{ \overrightarrow{F}_{weight} = -784.8 \text{ N} }\) |
\(\small\mathsf{ m = 80.0 \text{ kg} }\) \(\small\mathsf{ \overrightarrow{a} = 3.0 \text{ m/s}^2 }\) \(\small\mathsf{ \overrightarrow{F}_{weight} = -784.8 \text{ N} }\) |
| \(\small\mathsf{ F_{net} = m \overrightarrow{a} = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} }\) | \(\small\mathsf{ F_{net} = m \overrightarrow{a} = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} }\) |
| \(\small\mathsf{ (80.0 \text{ kg})(0.0 \text{ m/s}^2) = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ 0 \text{ N} = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = 785 \text{ N} }\) |
\(\small\mathsf{ (80.0 \text{ kg})(3.0 \text{ m/s}^2) = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ 240 \text{ N} = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = 1025 \text{ N} }\) |
| Downward Acceleration | Free Fall |
| \(\small\mathsf{ m = 80.0 \text{ kg} }\) \(\small\mathsf{ \overrightarrow{a} = -3.0 \text{ m/s}^2 }\) \(\small\mathsf{ \overrightarrow{F}_{weight} = -784.8 \text{ N} }\) |
\(\small\mathsf{ m = 80.0 \text{ kg} }\) \(\small\mathsf{ \overrightarrow{a} = -9.81 \text{ m/s}^2 }\) \(\small\mathsf{ \ \overrightarrow{F}_{weight} = -784.8 \text{ N} }\) |
| \(\small\mathsf{ F_{net} = m \overrightarrow{a} = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} }\) | \(\small\mathsf{ F_{net} = m \overrightarrow{a} = \overrightarrow{F}_{normal} + \overrightarrow{F}_{weight} }\) |
| \(\small\mathsf{ (80.0 \text{ kg})(-3.0 \text{ m/s}^2) = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ -240 \text{ N} = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = 545 \text{ N} }\) |
\(\small\mathsf{ (80.0 \text{ kg})(-9.81 \text{ m/s}^2) = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ -784.8 \text{ N} = \overrightarrow{F}_{normal} + -784.8 \text{ N} }\) \(\small\mathsf{ \overrightarrow{F}_{normal} = 0 \text{ N} }\) |
