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More on Normal force (shoe on floor)

David explains how to determine the normal force for a variety of scenarios (extra forces, diagonal forces, acceleration) involving a shoe on the floor. Created by David SantoPietro.

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  • starky ultimate style avatar for user Tim
    What about the instant when a falling shoe hits the ground? What would be the normal force then?
    (22 votes)
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  • duskpin ultimate style avatar for user Ankita Behera
    I saw a question: a boy of mass M_ holding a box of mass _m above his head jumps from a building. What is the force exterted by the box on his head during the freefall? Does it increase during the period he balances himself after hitting the ground?

    Is there a normal force when two bodies in contact are in freefall? Will they stay in contact?
    (18 votes)
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    • spunky sam red style avatar for user V_Keyd
      Great question! i don't want to give a straight up answer because you'd appreciate it more when you'll come to conclusion yourself. Let's think about what Galileo told us about object in a free fall. A stone and a feather, ignoring any air resistance, will fall from a building at the same rate! Of course, you knew that already. But it's the key to your answer. Imagine yourself holding a ball in your palm and jumping from a high diving board. Both you and ball are falling at the same rate. Which means if you removed your palm from underneath the ball, it would still keep falling the same way. Think another case: Even if you didn't have the ball on your palm and both you and the ball had been dropped side by side from the diving board, the picture would still be the same.

      So is there a normal force on the ball in free fall?

      If you don't get the answer, ask again. I promise to tell the answer but you have to give the problem your best effort. Read the chapter again.
      (20 votes)
  • female robot amelia style avatar for user McSmith Nwalozie
    what about the force on the horizontal direction? what happens to it?
    (11 votes)
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  • hopper cool style avatar for user Vishal
    Why is sin phi used instead of sin theta?
    (4 votes)
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  • duskpin ultimate style avatar for user elenast997
    In the 'shoe falling through the air' case, what if we take in consideration the air resistance? Could that be viewed as a normal force? (it still is a contact force, since we're talking about the air molecules reacting to the shoe in the opposite direction when they come in contact).
    Of course, in this case the normal force (air resistance) would be a lot smaller than the force that caused it (mg), but then again that's why it's falling through the air, not levitating.
    (10 votes)
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    • sneak peak green style avatar for user Jellyacejunior
      Seeing as how this was made 7 years ago, you might not care anymore but I would like to still give my answer.

      Air resistance like normal force is a kind of contact force, forces acting upon objects in contact. Air resistance is really a special kind of frictional force(also a contact force). The difference between frictional and normal force that you will see in diagrams is that the normal force is perpendicular to the place of contact while friction is along it or parallel.

      Normal forces are also described as the reactionary force of the compounds or atoms of a material to not allow another object to pass through it. This is an important distinction in comparison to frictional forces which does "let an object pass through it" or does not completely stop an object despite being in contact with it. For example gases(air resistance) and liquids(water resistance) and solids whose point of contact is parallel to the force acting upon it.

      However, as you've mentioned isn't this kind of the same fundamental force. And you are right! fundamentally, they are really the same forces, intermolecular ones at points of contact. However, because of the fact that macroscopically normal force, air resistance, water resistance, and the usual kind of friction are so different looking and whose emergent properties differ, it is much easier to deal with them as different kinds of forces.
      (2 votes)
  • mr pants green style avatar for user Manasv
    Hi, If acceleration = force/mass doesn't that mean that the object with more mass falls slower than object with less mass? now here doesn't mass means mass of the object as f=ma so we're taking force on the object. But that doesn't happen according to experiments. Where am I wrong?
    (3 votes)
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  • blobby green style avatar for user Rohan Jaswal
    In the above video what will happen to the horizontal force F3x ?
    (4 votes)
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  • spunky sam blue style avatar for user bilalquetta457
    When he gives the example of Fsub2 being applied in the upward direction,the equation will become
    Does this means that the weight of shoe will be increased? (I'm referring this concept from the weightlessness in an elevator)
    Please clearify this to me.I'm finding this very confusing.In case of an elevator it was obvious but here it is not making sense..
    (3 votes)
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    • piceratops ultimate style avatar for user Kartikeye
      No, the weight remains constant, remember weight cannot change, as long as you are near the surface of the Earth. Because remember, weight is just the measure of the force of gravity on an object. No, the weight of the shoe will not increase, the normal force will decrease ( assuming the acceleration is 0 ), because the F2 relieved some of the force that was needed to be applied by the normal force.
      I hope this helps.
      Comment if you still have some doubts.
      -ƙαɾƚιƙҽყҽ ʂԋɾιʋαʂƚαʋα™
      (4 votes)
  • piceratops tree style avatar for user Amrutha G
    At , why doesn't the net force on the RHS change if the total acceleration of the shoe changes? Won't there be an extra force due to this acceleration?
    (3 votes)
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    • piceratops sapling style avatar for user Aidan Loewen-Thomas
      Remember that acceleration is not actually a force; acceleration is proportional to the net force. No additional forces will be acting on the shoe. The only difference after you add an acceleration is what the net force is equal to. If you had a resting object, then Fnet = ma, but acceleration is 0, so Fnet = 0; if you had an accelerating object, then Fnet = ma, and since acceleration does not = 0, then you actually have to account for this in your calculation.
      (2 votes)
  • starky tree style avatar for user Rajan
    Doesn't newton say "Every reaction has an equal and opposite reaction"?doesn't the normal force represents to the opposite reaction?So, the normal force shouldn't be equal to the force is given?please clear me that.{When the surface is tilted}
    (1 vote)
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    • leaf green style avatar for user Alexander Wu
      Nope. Action and reaction forces operate on different objects, not the same object. The reaction force to the normal force the surface exerts on the object (up) is the normal force the object exerts on the surface (down). The reaction force of the gravity earth exerts on the object (down) is the gravity the object exerts on earth (up). The reason earth isn’t affected much by this gravity is that its mass is much much greater, so its acceleration is negligible.

      So to sum up, Action and reaction forces operate on different objects and are the same type of force (reaction force of normal force is still normal force, reaction force of gravity is still gravity, reaction force of friction is still friction). In addition to that is Newtons third law.
      (5 votes)

Video transcript

- [Voiceover] Check out this fine looking sneaker right here, we're gonna use this shoe to illustrate some more challenging normal force problems and we're gonna 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 the shoe's in contact with the floor, there will 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 would 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 make a mistake, they do this, they say alright, 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, whoop, there's a normal force there's gotta 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 this 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, and this is what people wanna 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 wanna 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, and that'll be true in the simple case, but I'll show you coming up how that's not gonna 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 that 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, and in this case, since these are vertical forces, I'm gonna consider the acceleration in the vertical direction and the net force in the vertical direction. And 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 it's motion, not changing it's velocity, the acceleration's just gonna be zero, so the vertical acceleration, acceleration, excuse me, should be zero. For the net force, I've got an upward normal force, so I'm gonna make that positive, if fn represents the magnitude of the normal force, this would be positive Fn, I'm just gonna put positive here even though I don't really need it but to show you that it's upward, we're gonna consider upward to be positive and then I've got this downward gravitational force, and if mg represents the size of the gravitational force, I'm gonna put a negative here to represent that that gravitational force is down, and 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'll have normal force minus mg, and if I finally solve for the normal force, I'll get that the normal force is gonna 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 because 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 gonna 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 gonna 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 gonna have another force, I'm gonna have F1, that points down so in my force diagram I'd have another force that points down, F1, that means I'd 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 gonna be bigger, bigger by an amount F2 and that makes sense, if you push down on a, oh no F2, wow, F1, sorry about that, it's gonna 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 gonna be greater, you're squashing these two surfaces together with greater force, so the ground's gotta 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 because they just, they love the shoe, they recognize the beauty of the shoe, well crafted shoe, so if there's an F2 pointing up, we now have another force in our diagram, that force would point up, we would call this F2, over here how would this change? Again, still acceleration to 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 I have to subtract F2 from both sides so now I'd 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 gonna reduce the normal force by 20 Newtons because 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 and you don't know what to do, let's say, we have another force, let's say this force is gonna be a diagonal force, so we're gonna pull this way. That was not a well drawn force, let me draw it like this. So we got a force this way at an angle. Now we're talking, 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 gonna 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 wanna 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 alright F3 is gonna have a vertical component, that vertical component, I'll call it F3 Y, for F3 in the vertical direction. And it's also gonna have a horizontal component. I'll just call that F3 X for F3 in the horizontal direction. So if I wanna solve for F3 Y, I'll just use the definition of sine and I know to use sine because this side is the opposite to this angle, I know sine relates opposite side, so I'm gonna write this as sine of phi is gonna equal the opposite side is F3 Y so F3 in the Y direction divided by F3 total, the total magnitude of F3, and if I solve this for F3 Y, I get F3 in the Y direction is gonna equal F3 times sine of Fi, now I can include this in my force, my net force, because this points upward, so since it points up, this vertical component's gonna add a plus, F3 sin theta over here, or sorry not theta, it's gonna be F3 sin Fi, and I'll have a plus F3 sin fi right here, and when we subtract this F3 sin phi from the other side to get it over to here, we're gonna get minus F3 sine phi 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 the F3 sin 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 A zero, in that case, nothing would change on this right hand side. Sometimes people think if there's acceleration, there's gonna 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 a zero 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 a m a zero, and then a plus m a zero when you solve for the normal force. Alright, I think we've pretty much exhausted this example of a shoe on the floor, it'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.