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How the normal force changes when an elevator accelerates. Created by Sal Khan.
Video transcript
What I want to do in this video is think about how the normal force might be different in different scenarios Since my 2.5-year-old son is obsessed with elevators I thought I would focus on those So here I've drawn 4 scenarios We can imagine them almost happening in some type of the sequence So in this first picture right over here I am going to assume that the velocity is equal to zero or another way to think about this is, the elevator is stationary And everything we're gonna be talking about in this video is the vertical direction That's the only dimension we're going to be dealing with So 0 m/s in the vertical dimension Another way to think about it, this thing is not moving Now it is also--this may be somewhat obvious to you, but its acceleration is also 0 m/s^2 in this picture right over here Then let's say that I'm sitting in this transparent elevator and I press the button So the elevator begins to accelerate upwards So in this screen right over here, let's say the acceleration is 2 m/s^2 I'll use a convention that positive means upwards or negative means downward So we're only going to be operating in this one dimension right here I could write 2 m/s times the j unit vector because that tells us where we're moving. I'll just live it like that That tells us we're moving in the upward direction Let's say we do that for one second and then we get to this screen right over here So we have no velocity; we accelerate-- oh, this is 2 m/s squared. This is acceleration here So we do that for one second and at the end of one second, we stop accelerating So here in this screen over here, our acceleration goes back to 0 m/s^2 in the j direction. That's really just zero But now we have some velocity Just for the sake of simplicity, let's say this screen lasted for 1 second So now our velocity's going to be to 2 m/s in the j direction or in the upward direction And then let's say we do that for 10 seconds So with that constant velocity, we travel for 20 meters We traveled a little bit when we were accelerating too When we're getting close to our floor the elevator needs to decelerate So it decelerates. The acceleration here is -2 m/s^2 times--in the j direction It's actually accelerating downward now. Has to slow it down to get it back to stationary So what I want to do is think about what would be the normal force the force that the floor of the elevator is exerting on me in each of these situations We're going to assume that we're operating near the surface of the earth So in every one of these situations, if we're operating near the surface of the earth I have some type of gravitational attraction to the earth and the earth has some type of gravitational attraction to me I'll just make the math simple--let's say that I am some type of a toddler and I'm 10 kg So maybe this is my son, although he's 12 kg, but we'll keep it simple So let me make it clear. He doesn't weigh 10 kg. That's wrong He has the mass of 10 kg; weight is the force due to the gravity mass is the amount of matter there is Although I haven't found the rigorous definition So the mass of the individual, of this toddler sitting at the elevator is 10 kg So what is the force of gravity, or another way to think about it, what is this person's weight? Well, in this picture right over here, it's mass times the gravitational field near the surface of the earth 9.8 m/s squared And the negative tells you it's going downwards So you multiply this times 10 kg The downward force, the force of gravity is going to be 10 kg times -9.8 m/s squared So -98 newtons. That's in the j direction Well, what's going to be the downward force of gravity here? It's going to be the same thing. We're still near the surface of the earth We're going to assume that the gravitational field is roughly constant although we know it slightly changes with the distance from the center of the earth but when we're dealing on the surface, we assume that it's roughly constant And so we'll assume we have the exact same force of gravity there and of course this toddler's mass does not change depending on going up a few floors So it's going to have the same force of gravity downwards in every one of these situations In this first situation right here the person has no acceleration If they have no acceleration in any direction and we're only concerning ourselves with the vertical direction right here that means that there must be no net force on them This is from Newton's first law of motion But if there's no net force on them, there must be some force that's counteracting this force because if there was nothing else, there would be a net force of gravity and this poor toddler would be plummeting to the center of the earth So that net force in this situation is the force of the floor of the elevator supporting the toddler So that force would be an equal force, but in the opposite direction and in this case that would be the normal force. So in this case the normal force is 98 N in the j direction So just completely balances off. There's no net force on this person they get to hold their constant velocity of zero; they don't plummet to the center of the earth Now what is the net force on this individual right over here? Well, this individual is accelerating; there is acceleration going on over here So there must be some type of net force Let's think about what the net force must be on this person, on this toddler I should say The net force is going to be the mass of this toddler, 10 kg times acceleration of this toddler, times 2 m/s squared which is equal to 20 kg m / s squared, which is the same thing as 20 N, 20 N upwards 20 N upwards is the net force So if we already have the force due to gravity at 98 N downwards That is the same thing here 90 N downwards We need a force that not only balances off that 90 N downwards not only keep stationary, but also doing another 20 N in the upward direction Here we need a force in order for the elevator to accelerate this toddler upwards at 2 m/s You have a net force of positive 20 N, or 20 N in the upward direction Or another way to think about it if you have -98 N here you're going to need 20 more than that in the positive direction So you're going to need 118 N, now in the j direction So here where the elevator is accelerating upward the normal force the is now 20 N higher than it was there and that's what's allowing this toddler to accelerate Now think about this situation No acceleration, but we do have velocity So here we were stationary; here we do have velocity And you might be tempted to think, oh, maybe I still have some higher force here because I'm moving upwards. I have some upwards velocity But remember Newton's first law of motion. If you had a constant velocity including a constant velocity of zero, you have no net force on you So this toddler right over here, the net forces is gonna look identical over here And actually if you're sitting in either this elevator or this elevator assuming it's not being bumped around at all you would not be able to tell the difference because there's no-- your body is sensitive to acceleration your body cannot sense velocity that has no frame of reference or nothing to see passing by So the toddler there doesn't know whether it is stationary or there's constant velocity It would be able tell this; it would feel that kind of compression on its body and that's what its nerves or sensitive perception is sensitive to But here it's identical to the 1st situation and Newton's first law tells us that there's no net force on this So it's just like the first situation the normal force, the force of the elevator on this toddler's shoes is going to be identical to the downward force due to gravity So the normal force here is going to be 98 N; completely nets out the downward the -98 N It's in the j direction, in the positive direction And then when we're about to get our floor, what is happening? Well, once again we have a net acceleration We have a net acceleration of -2 m/s^2 So once again what is the net force here? The net force over here is going to be the mass of the toddler 10 kg times -2 m/s^2 This is in the j direction, the vertical direction Remember j is just the unit vector in the vertical direction facing upwards So -2 m/s^2 in the j direction And this is equal to -20 kg m / s^2 in the j direction or -20 N in the j direction So the net force on this is -20 N So we have the force of gravity -98 N in the j direction So we we're not fully compensating for that because we're still gonna have a net negative force while this child is decelerating And that negative net force is -20 So we're only going to have a 78 N normal force here that counteracts all but 20 N of the force due to the gravity This right over here is going to be 78 N in the j direction I really want you to think about this next time you're sitting in the elevator The only time that you realize that something is going on is one that elevator's really just accelerating or when it's just decelerating When it's just accelerating, you feel a little bit heavier When it's just decelerating, you feel a little bit lighter I want you to think a little bit about why that is While it's moving at a constant velocity or stationary you feel like you're just sitting on the surface of the planet someplace