Woman in Elevator Problem (Free Fall), Forces she experiences?

In summary, the problem involves determining the impact force experienced by a woman in an elevator that free-fell 6 feet and abruptly stopped. The weight of the elevator is unknown and the woman weighs 140lbs. The equations used are x = x_0 + v_0*t + (1/2)a*t^2 and F = ma. In the worst case scenario, where all the weight is acting on one ankle and the elevator hits the ground and stops instantaneously, the ground reaction force is equal to the weight of the elevator plus 623.23N. However, this scenario is not physically accurate as it implies infinite force. In the best case scenario, where the body weight is equally distributed through both ankles and the elevator
  • #1
Ballin27
5
0
Hello, first post. I hope that someone can help me with this. I will roughly summarize the problem, and what I've done thus far.

Homework Statement


A lady was in an elevator that free-fell 6 feet and abruptly came to a halt. We had to determine a reasonable best case and a reasonable worst case value for the impact force that she experienced. The weight of the elevator is unknown and we are to find a value which I have not done yet. The woman weighs 140lbs.


Homework Equations


x = x_0 + v_0*t + (1/2)a*t^2
F= ma

The Attempt at a Solution


For the worst case I made the assumptions that all the weight was acting on one ankle and that the elevator hit the ground and came to a direct stop instantaneously.
The forces I had acting were her body weight, the weight of the elevator, and the ground reaction force.
I ended up with this:
F = 0 = -623.63 (N, bodyweight acting in the neg. direction) -Welevator(Weight of the elevator) + GRF (Ground reaction force)
Thus:
GRF = Welevator +623.23N
This was rather simple and once I find a weight of the elevator I should be fine.

My issue comes in the Best case scenario, where I assumed that the body weight was equally distributed through both ankles and that the elevator decelerates.
I had 1/2BodyWeight going through each ankle, the Welevator going through the middle, 1/2 of the GRF acting on each ankle.
I used:
x = xo + vo*t +1/2a*t^2
0 = 1.8288m + 0t + 1/2(-9.8m/s^2)*t^2
t = .61sec

Not sure if I can use this here, and my problem is deciding how I would get the force from there. I'm assuming I would need a vf and a vi as well as the mass of the elevator which I mentioned before. Would Impulse=momentum work here?

Any and all help would be greatly appreciated. Thanks in advance.
 
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  • #2
Ballin27 said:
Hello, first post. I hope that someone can help me with this. I will roughly summarize the problem, and what I've done thus far.

Homework Statement


A lady was in an elevator that free-fell 6 feet and abruptly came to a halt. We had to determine a reasonable best case and a reasonable worst case value for the impact force that she experienced. The weight of the elevator is unknown and we are to find a value which I have not done yet. The woman weighs 140lbs.


Homework Equations


x = x_0 + v_0*t + (1/2)a*t^2
F= ma

The Attempt at a Solution


For the worst case I made the assumptions that all the weight was acting on one ankle and that the elevator hit the ground and came to a direct stop instantaneously.
But an instantaneous stop would imply infinite force.
The forces I had acting were her body weight, the weight of the elevator, and the ground reaction force.
I ended up with this:
F = 0 = -623.63 (N, bodyweight acting in the neg. direction) -Welevator(Weight of the elevator) + GRF (Ground reaction force)
Thus:
GRF = Welevator +623.23N
This was rather simple and once I find a weight of the elevator I should be fine.
You can't apply Newton 1 when the body is decelerating. And leave the elevator weight out of this anyway. And please don't convert to Newtons when the problem is given in 'USA' units of feet and pounds :wink:
My issue comes in the Best case scenario, where I assumed that the body weight was equally distributed through both ankles and that the elevator decelerates.
I had 1/2BodyWeight going through each ankle, the Welevator going through the middle, 1/2 of the GRF acting on each ankle.
I used:
x = xo + vo*t +1/2a*t^2
0 = 1.8288m + 0t + 1/2(-9.8m/s^2)*t^2
t = .61sec

Not sure if I can use this here, and my problem is deciding how I would get the force from there. I'm assuming I would need a vf and a vi as well as the mass of the elevator which I mentioned before. Would Impulse=momentum work here?

Any and all help would be greatly appreciated. Thanks in advance.
You can't make any best case/worst case assumptions unless you know the elapsed time of the impact or distance through which the impulse force acts. The impulse force varies from near infinite in the worst case (falling onto a near rigid surface) to near zero in the other (falling into marshmallow fluff :yuck:). Welcome to PF!:smile:
 

1. What is the Woman in Elevator Problem (Free Fall)?

The Woman in Elevator Problem (Free Fall) is a thought experiment used to illustrate the concept of free fall and the forces experienced by objects in a falling elevator. It involves a woman standing on a weighing scale in an elevator that is dropping with constant acceleration due to gravity.

2. What forces does the woman experience in the falling elevator?

The woman experiences two main forces in the falling elevator: her weight and the normal force from the elevator. Her weight is the force of gravity pulling her downward, while the normal force is the reaction force from the elevator pushing her upward to counteract her weight.

3. Does the woman's weight change as the elevator falls?

Yes, the woman's weight does change as the elevator falls. Since weight is a force and is dependent on the acceleration due to gravity, her weight will increase as the elevator falls faster and decrease as the elevator slows down.

4. How does the normal force change as the elevator falls?

The normal force also changes as the elevator falls. Initially, when the elevator is at rest, the normal force is equal to the woman's weight. As the elevator begins to drop, the normal force decreases to less than her weight, and when the elevator reaches its maximum acceleration, the normal force becomes zero. As the elevator slows down, the normal force increases back to her weight.

5. Does air resistance affect the forces experienced by the woman in the falling elevator?

Air resistance does not significantly affect the forces experienced by the woman in the falling elevator. This is because the woman and the elevator are falling at the same rate, so the air resistance acting on them is equal and opposite, canceling out any effect on the forces experienced by the woman.

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