If an elevator.. was free falling at its max speed. Would i be able to jump while im in the elevator?
The max speed bit is irrelevant, free fall is free fall. Even though the normal reaction force is zero, it's still possible for you to launch yourself from the floor of the lift as long as it's within reach. Within the non-inertial frame of the falling lift, before you launch yourself, you are experiencing no force and you are stationary with respect to the lift floor. When you launch off, you are experiencing an upward force that accelerates you upwards relative to the floor and your velocity is upward away from the lift floor. Considering things from an inertial perspective outside the lift, you and the lift are both falling with an acceleration of g directed downward. There is no reaction force exerted by the lift floor on you. When you launch off, you are exerting a downward force on the lift that is accompanied by an upward reaction force by the lift on you. This upward force lessens your net downward acceleration, so that you momentarily experience less than g downward. Note that your velocity is still downward from the inertial perspective outside the lift.
i think that the answer is yes, but i also think that this is the case even before it reaches its max speed. perhaps you mean, are you still weigthless in the elevator also at the max speed? the maximum speed of the elevator i think is reached when it lands on the ground, and thus you'll not feel weightless. i may be mistaken, so wait for others to answer you.
Standing inside a falling elevator is the same as standing in a spaceship in space. If you can get your feet close enough to the ground then you can jump.
An orbiting shuttle is effectively a free-falling elevator. It's falling towards the earth, but it's also moving forward at 18000mph, so it falls "around" the curve of the earth and never gets any closer to the surface. The conditions inside the shuttle are basically be the same as inside your falling lift. You'd be weightless, and could push off the walls and ceiling etc, do pirouettes in the air, and fool around with globules of water. http://members.aol.com/petealway/orbit.html
You would not be able to tell the difference between an elevator at terminal velocity and an elevator that is stationary (until it hits the ground, of course...). You can jump exactly the same height in it in either case.
Well not unless you dropped your pen. You then might wonder why it is floating in the air in front of you. And when you jumped the fact you don't land would be a dead giveaway.
No, Art. If the elevator is at its terminal velocity, as the op and I stated, it would not be accelerating and therefore nothing would seem different from if it was standing still (or moving at a constant velocity, supported by its cable). Only when it is accelerating (at the start of the fall) would you be weightless.
The thread is entitled Free Falling Elevator. The second poster clarified that, it's terminal velocity is when it hits the ground at which time you're jam and won't be jumping anywhere.
Art, terminal velocity is the maximum speed that a given body will accelerate to in free fall. Given enough time or great enough height, that is reached well above ground level. The OP specified that the elevator is at its maximum speed, so that has been reached even though it's still falling.
Okay - I read it the same as poster 2 that in this thought experiment other factors were discounted and so the elevator would continue to accelerate at g until it landed.
The question is not well worded, so it isn't bad to deal with both cases. The first few posters assumed one case and I assumed the other.
I think it is best to wait for the OP to come back and explain him/herself. I have a feeling that this is another case of "bad wording" with a misconception of "free fall" implying "maximum speed" of some kind. Zz.
I agree. The OP's question was very badly written and shows more than a little confusion of the concepts.
The answer is yes, and no. I'll explain why. (This is a very VERY poorly written problem.) The elevator cable snaps, and you and the elevator go into free fall. As you and the elevator fall, the elevator will fall FASTER than you do, because it weighs MORE than you do. Because the elevator falls faster, there is the possiblity that it looses contact with your feet. case (1): The elevator separates from you before reaching terminal velocity. Here, you can jump *BEFORE* you separate from the floor. But if you dont, there's still hope! The elevator will reach a faster terminal velocity than you do because it weighs more, as already mentioned. BUT, as the elevator nears the bottom, it is going to compress the column of air that was below the elevator. This net pressure force will retard the motion of the elevator. IF it can slow the elevator down fast enough, you might be able to catch back up and remake contact with the floor (But perhaps a few split seconds before impact:surprised ). But then again, if the elevator is high enough, it could maintain its higher terminal velocity for a long enough duration that the distance between you and the elevator becomes so great that the dampening caused by the cushion of air below the elevator occurs before you ever catch back up. case (2): another possible case is that you never separate before reaching reaching terminal velocity with the elevator. This occurs if the elevator falls at a rate that is negligible as compared to your rate of descent. You can jump from the floor at *ANY* point along your fall.
Galileo established this already as not being true ! Big stones and small stones fall just as fast from the tower of Pisa. Things freely fall with an acceleration of 1 g (= 9.81 m/s^2) downward on the earth's surface. There are two ways to see this, one is the Newtonian way, and the other is the Einstein way. The Newtonian way is this: The WEIGHT (= force excerted by gravity upon an object at the surface of the earth) is given by m.g. Putting this force in Newton's equation gives you: m dv/dt = m.g The m cancel. We have dv/dt = g, no matter what's the weight. In fact, this is due to the equality of the two "m" in this problem. The m in m.g is called "passive gravitational mass", because it gives you the force as a function of the local gravity field. (the active gravitational mass would be the mass of the earth, which determines the value of g). It takes on the function that electric charge has in electromagnetism: the force on a charge is q. E with E the electric field. And the "m" in Newton's equation m dv/dt = F, is called "inertial mass". A priori there's no reason why these two coefficients should be equal, but they are. And this is called the (weak) equivalence principle. It was discovered by Galileo. Now, Einstein went a bit further, and said that there is an underlying reason why these two are equal. Einstein said that physics for a free falling observer is essentially the same as physics in free space, and that we simply make a mistake in thinking that, at the surface of the earth, we are in an "inertial frame". The true inertial frame is the free falling observer, and at the surface of the earth, we are accelerating upward. The problem with accelerating upward is that in a flat space, we should see other people somewhat further accelerate away from us. So to reconcile the idea of "accelerating upward" and "staying at the surface of the earth", Einstein postulated that spacetime is curved: then you can say that at the same time you're accelerating upward, and remain at constant distance from the center of the earth. The whole idea is then that gravity is not really a force, but a consequence of the curvature of spacetime. This idea, plus the idea that physics looks as in flat spacetime in a free falling frame, is the (strong) equivalence principle. So, from Einstein's viewpoint, the freely falling elevator will behave as in outer space. Everything will be floating... until of course air outside of the elevator (which is accelerating upward in this frame) will start to drag along the elevator (so that it is not freely fallling anymore but dragged along the "upward accelerating air"). In that case, it looks more and more as if you are in a rocket, until finally you reach stationary velocity wrt the air, and it looks as if you are in a rocket accelerating at 1 g upward (like you do when walking on the surface of the earth).
I was talking about falling in air (where heavier masses DO fall faster), which is what is happening when you drop an elevator in a shaft with a fluid contained below that elevator. There is no justification to even consider making an argument that the elevator is in a vacuum. Edit: Sorry for not stating that from the onset. Actually, if the elevator shaft is not very big then yeah, you could neglect the terminal velocity; but, if were talking about the sears tower, maybe that assumption is no longer valid, and now you DO have to consider the heavier elevator falling faster. Galileo, ... that guy was a crackpot :rofl:
But then your argument works backwards: the elevator is exposed to the air resistance, while you, inside, aren't (unless there's a big hole in the floor of the elevator, in which case the situation changes). So you fall faster than the elevator which is slowed down by the air flow, no ?
:rofl: ! I just thought about that last night while lying in bed. You beat me to posting it!! Yes, you are 100% correct, YOU will always fall faster than the elevator, so your feet will always keep contact with the floor of the elevator.