Why don't we "fly up" in an accelerating elevator?

In summary: If ##N > mg##, and ##m## doesn't move, it feels the same as if the surface didn't move and ##m## was "bouncing" upward.
  • #1
ago01
46
8
Earlier I was doing a sample problem for class that involved the work done by an elevator, and the problem gave us the normal force experienced by the person in the elevator (to calculate the acceleration of the elevator-person system).

I had done this wrong because I had wrongly assumed because the person isn't leaving the floor of the elevator that the net force from N and gravity must balance. After doing the calculations, it was clear N > mg, so there is an acceleration.

Mathematically I can work this out and if I had to regurgitate something on an exam I could do it just from the force diagrams. But I am wondering what the physical manifestation is.

What (I think) I do understand is the concept of the perceived weight of the person. If we stand them on a scale while the elevator is accelerating upwards, they feel heavier. If we do the same as the elevator is slowing down (or going down) they feel lighter. This is also consistent with my experiences. I suppose the problem I'm having is that if there is an imbalance between a surface and the normal force I had this idea that the object would simply fall through that surface (if N < mg) or "bounce" off the surface (if N > mg) and the only reason an object is ever held stationary on a surface is because the normal force exactly balances the gravitational force. This was reinforced by problems that, for example, drop a ball off a cliff. There's no normal force, so the downward acceleration is provided entirely by the gravitational force and the object falls. Or alternatively, a sliding block where it's not leaving the surface because N = mg.

In this elevator though the person isn't being "launched" up as the elevator accelerates. At least this hasn't been my personal experience and I've survived many elevator rides. Also, N > mg. So it seems to me that when moving vertically with a surface the normal force is the force providing the acceleration and no extra forces are required. Then, it would also seem that if I hit a ball with a paddle it's the normal force of the paddle's surface providing the acceleration to the ball at the time of impact (even though we would likely just simply treat this as another force). Is this the right idea?
 
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  • #2
We could "fly up" if the elevator accelerates our bodies and then slow down or stop, just like the paddle does respect to the ball.
We could "sink through the floor" if the elevator allows us to fall and then slow down or stop, if that floor were not a solid surface.
 
  • #3
Lnewqban said:
We could "fly up" if the elevator accelerates our bodies and then slow down or stop, just like the paddle does respect to the ball.

So I suppose then as long as the surface is moving at the same acceleration relative to us we will move with that surface and the normal force provides the acceleration. Your explanation also makes sense, if the elevator suddenly stopped the person in the elevator would keep going due to the larger normal force that was imposed on it the instant before it stopped.
 
  • #4
ago01 said:
if the elevator suddenly stopped the person in the elevator would keep going due to the larger normal force that was imposed on it the instant before it stopped.
If the elevator suddenly stopped, the person would keep moving by Newton's first law. They would be in projectile motion under gravity until they hit the ceiling of the elevator.

I don't understand your ideas about normal forces. A bat hitting a ball projects the ball because the bat accelerated to a significant non-zero velocity before impact. An elevator starts accelerating those inside it as soon as it starts moving, so there is no sudden impact, but a common acceleration.
 
  • #6
ago01 said:
if there is an imbalance between a surface and the normal force I had this idea that the object would simply fall through that surface (if N < mg) or "bounce" off the surface (if N > mg)
It is exactly the opposite:
  • If ##N > mg##, and ##m## doesn't move, either the surface is breaking apart and ##m## is going through it (feeling like "falling" through it), or ##m## is deforming or breaking apart;
  • IF ##N < mg##, and ##m## doesn't move, it feels the same as if the surface didn't move and ##m## was "bouncing" upward.
 
  • #7
jack action said:
It is exactly the opposite:
  • If ##N > mg##, and ##m## doesn't move
Then Newton's second law is being violated.
 
  • #8
jbriggs444 said:
Then Newton's second law is being violated.
Not if elasticity is involved, or something breaks.
 
  • #9
jack action said:
Not if elasticity is involved, or something breaks.
If net force is non-zero and acceleration is zero, that's a problem.
 
  • #10
If the elevator's acceleration exceeds ##g## then indeed objects inside will fly up towards the ceiling, however below ##g## objects inside are still bound to the floor of the elevator by gravity.
 
  • #11
In the stationary ground frame of reference, both the elevator and the man have the same acceleration, and there is no relative acceleration between them. In the non-inertail frame of reference of the elevator, neither the man nor the elevator appears to be accelerating, but there is an additional fictitious gravitational acceleration (and force) on both of them, but again no relative acceleration. So, in the non-inertail frame of reference of the elevator, the total normal force on the man is N=mg', where g' = g + a.
 
  • #12
Another example would be the vomit comet:

 

FAQ: Why don't we "fly up" in an accelerating elevator?

What is the concept behind "flying up" in an accelerating elevator?

The concept behind "flying up" in an accelerating elevator is based on Newton's First Law of Motion, which states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force. In an accelerating elevator, the external force is the elevator's movement, causing the passengers to feel as though they are flying up.

Why do we feel weightless in an accelerating elevator?

We feel weightless in an accelerating elevator because the force of gravity acting on our bodies is equal to the force of the elevator's acceleration. This creates a sensation of weightlessness, as our bodies are not experiencing a net force in any direction.

Will we continue to "fly up" forever in an accelerating elevator?

No, we will not continue to "fly up" forever in an accelerating elevator. Eventually, the elevator will reach a constant velocity and the sensation of flying up will cease. This is due to the elevator's acceleration becoming equal to the force of gravity, resulting in a net force of zero on the passengers.

How does the speed of the elevator affect the sensation of "flying up"?

The speed of the elevator does not affect the sensation of "flying up." As long as the elevator is accelerating, the passengers will continue to feel weightless and as though they are flying up, regardless of the speed at which the elevator is moving.

Can we experience the same sensation of "flying up" in a non-accelerating elevator?

No, we cannot experience the same sensation of "flying up" in a non-accelerating elevator. In a non-accelerating elevator, the force of gravity is the only force acting on our bodies, resulting in a feeling of weight and being stationary. Without an external force, we cannot experience the sensation of flying up.

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