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

Click For Summary
SUMMARY

The discussion revolves around calculating the impact force experienced by a woman in a free-falling elevator that descends 6 feet before stopping. The woman weighs 140 lbs, and the weight of the elevator is unknown. The worst-case scenario assumes all weight is concentrated on one ankle during an instantaneous stop, leading to a ground reaction force (GRF) equation of GRF = Welevator + 623.23 N. In the best-case scenario, the weight is distributed across both ankles, and the elevator decelerates, requiring additional calculations for time and impulse to determine the forces involved.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly F=ma.
  • Familiarity with kinematic equations, specifically x = x_0 + v_0*t + (1/2)a*t^2.
  • Knowledge of impulse and momentum concepts in physics.
  • Basic understanding of forces and weight distribution in physical scenarios.
NEXT STEPS
  • Research how to calculate ground reaction forces in dynamic scenarios.
  • Learn about impulse-momentum theorem applications in real-world situations.
  • Study the effects of weight distribution on impact forces in free fall scenarios.
  • Explore the implications of instantaneous stops in physics and their mathematical representations.
USEFUL FOR

Physics students, engineers, and anyone interested in mechanics, particularly those studying forces in dynamic systems and impact analysis.

Ballin27
Messages
5
Reaction score
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.
 
Physics news on Phys.org
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 ). Welcome to PF!:smile:
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
12
Views
4K
Replies
17
Views
4K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K