Transfer of Force?

Question

If you were in an elevator, 400ft high, and the cable snapped, if you jumped just before the ground collision, would you be ok? (Assuming the ceiling is 20ft so you have no chance of hitting head)?

The way I think of it is that you transfer your force to the elevator.

Answer 1

The thing is you cannot jump in a free-falling elevator. Say the elevator is free falling, bear in mind that you are free falling with the elevator at the same time. Therefore there's no force between your feet and the elevator. You and the elevator are merely two objects falling at the same time. Since you cannot exert a downward to the elevator floor, the elevator cannot exert an upward force for you to jump up. Therefore, when the elevator lands on ground, it stops and you'll be stopped by the elevator floor. The kinetic energy is dissipated by the crumbling of the metal body and the kinetic energy of your body is dissipated by internal breakdown.

Answer 2

From F = p/t = m sqrt(2gh)//H/sqrt(2gh); where m = your mass, h = 400 ft fall, and H = 20 foot ceiling, we can find F = 40 mg = 40 W. W = mg which is your weight. In other words, without jumping, you'll smack into the ground with a force that is 40 times your weight.

t = H/sqrt(2gh) is the assumed time of impact because that's how long it will take the elevator of height H to collapse at velocity v = sqrt(2gh), the impact velocity.

Anyway if you weigh in at a buff 200 pounds, you'll smack the floor of the collapsing elevator with a force of 8,000 pounds (4 tons). And that could hurt. So you try the old, jump before impact bit.

How high can you jump vertically? 4 ft at most I'll bet. That means, relative to the elevator's floor, you need a kinetic energy KE = 1/2 mV^2 = mgs = PE to convert to potential energy at s = 4 ft above the elevator floor. Then V = sqrt(2gs) ~ 16 fps jump velocity. (g ~ 32 ft/sec^2).

Assume you need to accelerate a >= V/t = sqrt(2gs)//H/sqrt(2gh), where t is the same as before to get to speed V for the four foot jump. Thus a >= sqrt(2gh) sqrt(2gs)/H = 2g sqrt(hs)/H Then the force needed to jump four feet is f = ma = 2mg sqrt(hs)/H ~ 2mg 40/20 = 4 mg or four times your weight.

[Even when buffed, it's unlikely you can push off with a force of 800 pounds. A more modest s = 2 ft jump would require a force f = 2.8 mg or 560 pounds push. Why? Because t = H/sqrt(2gh) = 20/160 = 1/8 second...a much shorter time than the usual jump time. This is yet another reason jumping to relieve the impact force is an unlikely option. There is very little time to jump in.

In contrast, if you had 1 second to jump in, then f = ma = m sqrt(2gs)/1 = 16 m = 1/2 mg or half the body weight in force (e.g., 100 lbs) would do the job for a four foot jump. But then f would be much smaller; so the lessening of the impact would be much smaller as well.]

In other words, your impact F' = F - f = 36 mg...36 times your own weight when you jump up four feet off the floor...if you could make the jump in the 1/8 second time alloted. Hey, now you'll be crushed by 3.6 tons of force instead of 4 tons. Sorry, that could still give you a serious headache.

Answer 3

lets ignore some physics and assume that the person was falling with the lift and he's able to push the floor of the lift.

by the time the elevator reaches the ground, the velocity that the person is travelling would be already very high. hence, he would need to exert a tremendous amount of force on the elevator's floor to change his velocity to upwards and jump off the lift.

i would estimate that he would need to generate about 2900N of force (by calculations) in lets say a second so that he can jump off the lift, which is almost impossible.

hence, he most probably would not be alright. =X

Answer 4

If you fell 400 ft in an elevator, by the time you reached the ground you would not be on the floor anymore, so you wouldn't be able to jump. You wouldn't be okay if you were able to jump anyways, there's too much force from a fall of 400 ft.