More on Normal force (shoe on floor) | Physics | Khan Academy

Check out this fine looking sneaker right here. We're going to use this shoe to illustrate some more challenging normal force problems, and we're going to take this as an opportunity to discuss a lot of the misconceptions that people have about the normal force.

So, one misconception is that people forget normal force is a contact force. You only have a normal force when two surfaces are in contact. So, when this shoe's in contact with the floor, there'll be a normal force on the shoe and a normal force on the floor. Or, if the shoe were in contact with the wall, there'd be a normal force on the wall and a normal force on the shoe.

But if the shoe were just falling through the air, here's what happens. For a lot of people, let's say the shoe's just falling, and you got a question, and the question said, "Draw the forces that are exerted on the shoe while it's falling through the air." People get so used to having normal forces that they do, they make a mistake. They do this: they say, "All right, let me draw it over here." They say there's a gravitational force, and that's just fine. There will be a gravitational force. There's always gravity. The Earth's always pulling down, and it pulls down with an amount mg.

But they're so used to having normal forces. I mean, normal forces pop up in so many different questions, it's almost just like a reflex. People just automatically put, "Ooh, there's a normal force! There's got to be a normal force, right? There's always a normal force!" But there isn't always a normal force. If the shoe is not in contact with the surface, you don't have a normal force. It's not until this shoe makes it to the ground or touches another surface that you'll have that normal force.

So if we stick the shoe right here and we let it rest on the ground, now you'll have a normal force, and that normal force will point up. This is what people want to say, and it's true when the surfaces are in contact. But if they're not in contact, you don't have a normal force.

And then here's another misconception: people think the normal force is always equal to mg. Because again, it's equal to mg in so many different scenarios that people just want to say, "Well, it's always equal to mg." And again, it's just like a reaction. People see normal force; they just automatically replace it with mg. That'll be true in the simple case, but I'll show you coming up how that's not going to be true and what you do if it's not true.

So, for instance, if we wanted to find what's the normal force if this shoe has a mass m. So let's assume the shoe has a mass m. What would the normal force be? We can use Newton's Second Law. We can always use Newton's Second Law.

So we'll say that acceleration equals the net force divided by the mass. In this case, since these are vertical forces, I'm going to consider the acceleration in the vertical direction and the net force in the vertical direction. So what is the acceleration for the shoe vertically if it's just sitting here in a room, sitting on the ground at rest and not changing its motion, not changing its velocity? The acceleration is just going to be zero.

So the vertical acceleration, excuse me, should be zero. For the net force, I've got an upward normal force, so I'm going to make that positive. If FN represents the magnitude of the normal force, this would be positive FN. I'm just going to put positive here, even though I don't really need it, but to show you that it's upward, we're going to consider upward to be positive.

And then I've got this downward gravitational force. If mg represents the size of the gravitational force, I'm going to put a negative here to represent that that gravitational force is down. Then I divide by the mass of the shoe, and if I do this, I get that these two forces, this net force divided by the mass, has to be zero according to Newton's Second Law.

But I can multiply both sides by the mass, and if I do that, the left-hand side is still zero, and I'll get that this is equal to the normal force minus mg. So I have normal force minus mg, and if I finally solve for the normal force, I'll get that the normal force is going to equal mg.

And a lot of people are like, "Yeah, I already knew that. Duh! Normal force is always equal to mg." But it's only equal to mg in this case, 'cause those were the only two forces. Look at the assumptions we made! Only two forces were the normal force and the gravitational force, and we assumed that the acceleration was zero. If you relax any of those requirements, normal force is no longer going to be equal to mg.

And it was on a horizontal surface. If you relax that requirement, again, there's no reason to think this has to be in the y direction. You could have normal forces in the x-direction.

So let's slowly, one point at a time, try to relax some of these requirements and see what that does to the normal force. In other words, what if we just added another force? What if we let the shoe sit here on the ground, and I push down on it? So I'm pushing down on this shoe. I'm going to say I'm pushing down with a force I'll just call it F1, so a force of magnitude F1, and it's pointing downward.

How would that change this now? So this is, we're stepping it up. This is going to be a little harder. What do we do? Well, the acceleration is still zero. Let's say it's still just sitting there, so we don't have to do anything with the left hand side; that's still zero. Multiplying by m still makes that zero.

But now up here in this force, up here, I'm going to have another force. I'm going to have F1 that points down. So in my force diagram, I'd have another force that points down, F1. That means I have to subtract it when I find the net vertical force. I'd have F1; this would be a negative F1 right here.

And when I solve for FN, I'd add mg to both sides to cancel it, and then I'd add F1 to both sides to cancel this F1, this negative F1, and I'd get mg plus F1. So I get the normal force is going to be bigger—bigger by an amount F1.

So that makes sense. If you push down on, oh not F2, wow, F1! Sorry about that. It's going to be bigger by an amount F1. So if I push down with an extra 10 newtons of force, there's more pressure, right? That makes sense. The pressure between the ground and the shoe is going to be greater. You're squashing these two surfaces together with greater force.

So the ground's got to push up to keep the shoe out of the surface. That's what this normal force does; it exerts a force to keep the object out of the surface, to keep the object from penetrating that surface. So if I push down on an object into a surface, that normal force increases, and it increases by the amount you're pushing down.

So that makes sense. If you had an upward force, let's say you had an upward force, someone's pulling up on the shoe while you push down. You're fighting over the shoe. You're wrestling over it with somebody 'cause they just, they love this shoe; they recognize the beauty of this shoe. Well-crafted shoe!

So if there's an F2 pointing up, we'd now have another force in our diagram; that force would point up. We'd call this F2 over here. How would this change? Again, still acceleration is zero. But I'd have an upward force now, so I'd have to add F2 vertically because that's another force. I'd have a plus F2 right here, and then over here when I solve for this, I add mg to both sides, I add F1 to both sides, and then I have to subtract F2 from both sides.

So now I would have FN is mg plus F1 minus F2. This also makes sense! If you pull up on a shoe, you're relieving some of the pressure between the shoe and the other surface, the shoe and the floor. So if I pull up with 20 newtons, I'm going to reduce the normal force by 20 newtons 'cause I'm relieving some of that pressure between the shoe and the floor.

Let's make it even harder! Let's make this thing scary. Sometimes you get really crazy problems; you don't know what to do. Let's say we have another force. Let's say this force is going to be a diagonal force, so we're going to pull this way. Oh, that was not a well-drawn force. Let me draw it like this.

So we got a force this way at an angle. Now, now we're talking! Let's say this is F3, F3 at an angle of, we'll call it fi. So the angle from this horizontal line here is fi. Now, what do we do? So I've got this crooked angle in here now. This F3 is going to be pointing this way, so I'll add another force to my force diagram, and I can figure out how to include this into my vertical force version of Newton's Second Law.

I can't include the entire F3 force. Here's a mistake people make. They want to just add F3 or subtract F3, but I can't do that. This is the vertical form of Newton's Second Law. This is only applying to the vertical direction, the y-direction. But F3 is pointing both vertically and horizontally!

So I have to only include the vertical part of F3 in this formula. So what I have to do is say that, all right, F3 is going to have a vertical component. That vertical component I'll call it F3y for F3 in the vertical direction, and it's also going to have a horizontal component. I'll just call that F3x for F3 in the horizontal direction.

So if I want to solve for F3y, I'll just use the definition of sine, and I know to use sine because this side is the opposite to this angle, and I know sine relates opposite side. So I'm going to write this as sine of fi is going to equal the opposite side is F3y, so F3 in the y-direction divided by F3 total, the total magnitude of F3.

And if I solve this for F3y, I get F3 in the y-direction is going to equal F3 * sine of fi. Now I can include this in my force, my net force 'cause this points upward. So since it points up, this vertical component is going to add a plus F3 sine fi over here.

Or, sorry, not fi; it's going to be F3 sine fi, and I'll have a plus F3 sine fi right here, and when we subtract this F3 sine fi from the other side to get it over to here, we're going to get minus F3 sine fi. And that makes sense because if we pull up, we know we're reducing some of the pressure.

We’ll reduce the normal force, so this component—look at this component—points up. It's reducing some of the pressure on the bottom of the shoe; the normal force goes down, and so we subtract F3 sine fi.

And one more way to step this problem up to the next level would be to say that this room isn't really just a room; maybe it's an elevator, and this elevator is accelerating upward with some acceleration az. In that case, nothing would change on this right-hand side. Sometimes people think if there's acceleration, there's going to be some new force, but if these are the forces, those are the forces.

The only thing that changes over here if this was in an elevator that's accelerating up is that instead of zero, you'd replace this with az or whatever the acceleration is. That's it! That's the only change! You could still solve for FN the same way when you multiply by m. It wouldn't be zero on the left anymore; you'd have an ma, and then a plus ma when you solve for the normal force.

All right! I think we've pretty much exhausted this example of a shoe on the floor. That's probably harder than any example you'll see, but now you know how to handle any type of force you might meet or acceleration.