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## Class 11 Physics (India)

### Unit 9: Lesson 5

Normal force and contact force- Normal force and contact force
- Normal force in an elevator
- Breaking down forces for free body diagrams
- Free body diagram with angled forces: worked example
- Normal force (basic)
- Forces at an angle
- Angled forces review
- More on Normal force (shoe on floor)
- More on Normal force (shoe on wall)
- What is normal force?

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# More on Normal force (shoe on wall)

David shows how to determine the normal force for a shoe shoved against a wall with a diagonal force. Created by David SantoPietro.

## Want to join the conversation?

- when i draw my free body diagram i am havng trouble with the trig of phi, i tend to always assume cos(phi) in the horizontal, is there a trick as to how you decided to use sin? I see how you would use sin in the diagram on the left but i cannot see where the corresponding phi angle would be on your free body diagram. thanx in advance(6 votes)
- Look at the trig from the perspective of the angle you are using. So, say you have a right triangle, and phi is on the top of that right triangle. The side opposite that angle is the sine, and the side of the triangle adjacent to that angle is cosine. I came up with word-play to remember. Cosine has "co" in front of sine, sort of like co-worker, so it is always going to be the side adjacent to the gang;e you are using in your trig calculations. Sine will always be the side opposite the angle, regardless is that's in the horizontal or vertical direction. A triangle is a triangle, it doesn't have direction itself. So the direction (vertical or horizontal) of the sides has no bearing on the trig calculations.(3 votes)

- I get this whole thing but i don't figure out why should we use F4Sin not F4cos(4 votes)
- Nope. Did you see his diagram? The angle that was given to him is opposite of the horizontal components. Soh Cah Toa Cosine is the side adjacent to the angle divided by the hypotenuse. Sine is the side opposite to the angle divide by the hypotenuse. Because the horizontal component is opposite of the given angle, you have to use sine. Hope this helps!(4 votes)

- then how are nails penetrated in walls or any solid(3 votes)
- The nail imparts more force than the strength of the cohesive forces holding the material together allowing the nail penetrate the material.(4 votes)

- Can there be more than one Normal Force on an object? (e.g. if the shoe is being pushed into the left bottom corner of the box)(4 votes)
- Why didn't he take the gravitational force into consideration while using Newton's second law?(2 votes)
- Because he is only taking into consideration the horizontal forces. ΣF = m⋅a is true for any direction. As the gravitational force occurs in the vertical direction, he just ommited it. That's why he only took into account the horizontal component of F4 (sin ϕ).(4 votes)

- If you're pushing against a surface like a sponge, what happens to the normal force?(3 votes)
- As you first touch the sponge, you are not exerting much force (and the sponge is not pushing back with much force), because the sponge is compressing and moving away from your finger. Once you have fully compressed the sponge, it will continue to push back as hard as you push.

I think there is more going on, because you are causing the sponge to compress, but it doesn't affect calculating the normal force.

I am not an expert. This is my understanding so far. :-)(1 vote)

- Just out of curiosity, when I'm pushing something like a shoe against a vertical wall extremely hard, the shoe doesn't fall to the ground although it seems like I only apply horizontal force to it. Is this some phenomenon that I don't know or just an illusion (maybe in reality I apply some vertical force to counteract gravity without realizing it)(2 votes)
- Because of the the force you are applying on the shoe it has a force between it and the wall. Because of the force between the wall and the shoe there is a static friction that is proportional to the force and the coefficient of static friction. This static friction is what resists the downward for frm gravity.(3 votes)

- I think I understand this, but what happens to the normal force if you break through a surface? For example, when you fall through a table.(3 votes)
- I was wondering, why are you using phi instead of theta? I've never worked with phi before, only theta.(1 vote)
- You can use any variable you want as long as you define and use it consistently.(4 votes)

- If a shoe was shoved into a corner of a box, would there be multiple normal forces or would we just say that there is one normal force which is diagonal?(3 votes)
- You need to observe the contact points between your shoe and the surfaces really well to know the answer. Nevertheless, the general answer will be three forces, one for each surface. If the end of your shoe get very close to the edge(and I'm assuming it is a very narrow shoe) the three surfaces will bend into a single surface. I think this is hard to imagine, but if you look at the 2D case its easier. If you have 2 lines that intersect at 90 degrees, at a real life situation the edge will be a very small curve.(0 votes)

## Video transcript

- [Narrator] Let's say we take this shoe. Instead of sitting it on the floor, here's one that trips people out. Let's say we take this shoe, and we shove it against the wall. So, walls can exert normal
forces just like floors can, but when this happens, people start to get a
little bit concerned, it starts to get a little bit weird. Let's say we exert a force, so say the force looks like this. So here we go, let's call this force F4. So here's F4, this force keeps
the shoe from falling down, but it also pushes the shoe into the wall, so again, we're gonna have a normal force, let me give you an angle here, let's say this angle right there is phi. And let's say the question we wanna ask, now, we wanna know what's the
normal force in this case? So this one's a little bit weirder, but we can still do it the same way, we should draw a force diagram first, it's always good practice, draw what forces are exerted on the object you're trying to find a force for. So, we're gonna have the normal force, but first we should draw
the force of gravity. Gravity's easy, gravity
always points down. So, you got mg straight down. We're gonna have a normal force, here's where people make a mistake. We will not draw the normal force up. People think that the
normal force is always mg, we saw that that's not true. People also think the
normal force is always up, but it's not. It's usually up because it's in contact with a horizontal surface. But now this is contact
with a vertical surface. And this word "normal" in
the phrase "Normal Force" is not referring to
like, boring, or usual, it's referring to "normal" in the mathematical
sense as perpendicular, perpendicular to the surface
exerting this normal force. And this wall, that's vertical, perpendicular to that wall
is coming out of the wall, and that's gonna be to the right, so the wall is gonna push
to the right on the shoe to keep the shoe from
penetrating this wall. So that's a little bit weird for people, is that this normal force
is now pushing to the right. Now I've got one more force, I've got my F4, so I'm
gonna draw this force, at 4 it looks something like that. Okay, so these are my forces. That's it, those are the
only forces there are. I mean, we're gonna neglect any friction, let's just assume that the
shoe's just sitting there, there's no other frictional forces, let's say this is it, we wanna find the normal force, what do we do? Again, we're gonna use
Newton's Second Law, we're gonna use a equals the net force in a certain direction, this time we're gonna use
the horizontal direction, we're gonna use the horizontal direction because the force we wanna
find, our normal force, is in the horizontal direction. So, the acceleration in the
x direction is gonna be what? Well, you think about this, if I'm pushing the shoe into the wall, it's probably got no
horizontal acceleration, even if it was sliding up and down. Even if there was motion up and down, it's probably not
penetrating into this wall and it's probably not
bouncing off of this wall, it's probably constricted to be only in the plain of this wall, so there's gonna be no
horizontal acceleration. And if that doesn't make sense, it's because there's no motion in the horizontal
direction, left or right. There's no velocity, change, at all, in this horizontal direction, because the shoe's not gonna be moving in that horizontal direction, and it continues to not move
in that horizontal direction. So our acceleration horizontally is just zero equals the net force, divided by the mass. Alright, the net force in the x direction. What are we gonna have in the x direction? Well I've got fn pointing to the right, so again that's a positive force I'm gonna consider
rightward to be positive, and I've got this F4, part
of it points to the left, so just like before, I've
gotta break this force up, I've gotta figure out how much of this force points horizontal, and how much of this
force points vertical, to get this F4 in the x direction, which is what I plug into
this formula up here, because I need this component here. This is the horizontal force of F4, not the vertical force. I don't plug the vertical
force in anymore, because this vertical force is
not part of the x direction, we're considering Newton's
Second Law for the x direction, so to solve for F4x, I'm
just gonna again use sine, because this angle, the
opposite of this angle is F4x. I'm gonna use sine of theta,
oh sorry, sine of phi. I'm gonna take sine of phi, that's gonna equal F4 and the x, divided by the total amount, F4, I get F4 in the x, is gonna
be F4 times sine of phi, and now I can use this up here, but you gotta be careful
with sines F4x points left, I'm gonna consider that a negative force. So if F4 sine theta
represents the magnitude, I'll write this as negative F4, sine, phi, sorry, I keep saying theta, I mean phi, I multiply both sides by m, I'll get zero again on the left hand side, equals, I've got Fn minus F4, sine phi, and now, when I solve this
for Fn, the normal force, I'll get the Fn, I'll add this
F4 sine phi to both sides, and I'll get that this normal force is gonna equal F4 sine phi. And that makes sense. It makes sense because, what these surfaces are doing, the reason why you're
getting a normal force, is these surfaces are exerting
whatever force they have to, to prevent any penetration
of this surface. So, if this F4x is pushing
in to the surface with F4x, right, if that's the force
we're pushing in with, Fn's just gotta equal that. It's gotta match that so that there's no acceleration horizontally. There were no other forces. We could, now you know
what to do if there were, if you wanted to step this up, you could add another force here, we'll call that F5, that'd
be another force this way, we'd have another F5, you
know how to handle that, now you'd come over to here, that's pointing to the
left, so you do minus F5, you'd come down here,
this would be a minus F5, you'd add that to both sides, that'd be a plus F5. What if we added a vertical force? What if we added another
vertical force this way to the shoe and we called that F6? Well that wouldn't impact
the normal force at all. This force F6 does not affect
how much these surfaces are getting pushed into each other. So I wouldn't include
that over here at all. That's a vertical force, it wouldn't affect the
normal force this time. Also note, gravity's not even affecting the normal force this time. 'Cause gravity's exerting a
force in the vertical direction, and our normal force is in
the horizontal direction. So, long story short, normal
force is not always mg, the normal force will only
exist, it'll only be non zero when two surfaces are in contact
and pushing on each other. You can change what the normal force is by adding forces into
or out of the surface, exerted on that object. And if there's a force at an angle, when you're finding normal force, make sure you only use the component that's in the same direction
as the normal force, 'cause that's the only
one that's gonna affect the normal force when you solve
using Newton's Second Law.