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- [Narrator] Let's say we take this shoe.

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Instead of sitting it on the floor,

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here's one that trips people out.

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Let's say we take this shoe,

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and we shove it against the wall.

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So, walls can exert normal
forces just like floors can,

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but when this happens,

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people start to get a
little bit concerned,

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it starts to get a little bit weird.

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Let's say we exert a force,

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so say the force looks like this.

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So here we go, let's call this force F4.

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So here's F4, this force keeps
the shoe from falling down,

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but it also pushes the shoe into the wall,

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so again, we're gonna have a normal force,

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let me give you an angle here,

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let's say this angle right there is phi.

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And let's say the question we wanna ask,

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now, we wanna know what's the
normal force in this case?

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So this one's a little bit weirder,

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but we can still do it the same way,

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we should draw a force diagram first,

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it's always good practice,

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draw what forces are exerted on the object

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you're trying to find a force for.

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So, we're gonna have the normal force,

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but first we should draw
the force of gravity.

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Gravity's easy, gravity
always points down.

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So, you got mg straight down.

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We're gonna have a normal force,

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here's where people make a mistake.

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We will not draw the normal force up.

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People think that the
normal force is always mg,

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we saw that that's not true.

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People also think the
normal force is always up,

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but it's not.

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It's usually up because it's in contact

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with a horizontal surface.

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But now this is contact
with a vertical surface.

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And this word "normal" in
the phrase "Normal Force"

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is not referring to
like, boring, or usual,

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it's referring to "normal"

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in the mathematical
sense as perpendicular,

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perpendicular to the surface
exerting this normal force.

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And this wall, that's vertical,

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perpendicular to that wall
is coming out of the wall,

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and that's gonna be to the right,

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so the wall is gonna push
to the right on the shoe

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to keep the shoe from
penetrating this wall.

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So that's a little bit weird for people,

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is that this normal force
is now pushing to the right.

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Now I've got one more force,

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I've got my F4, so I'm
gonna draw this force,

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at 4 it looks something like that.

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Okay, so these are my forces.

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That's it, those are the
only forces there are.

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I mean, we're gonna neglect any friction,

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let's just assume that the
shoe's just sitting there,

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there's no other frictional forces,

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let's say this is it,

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we wanna find the normal force,

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what do we do?

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Again, we're gonna use
Newton's Second Law,

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we're gonna use a equals the net force

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in a certain direction,

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this time we're gonna use
the horizontal direction,

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we're gonna use the horizontal direction

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because the force we wanna
find, our normal force,

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is in the horizontal direction.

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So, the acceleration in the
x direction is gonna be what?

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Well, you think about this,

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if I'm pushing the shoe into the wall,

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it's probably got no
horizontal acceleration,

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even if it was sliding up and down.

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Even if there was motion up and down,

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it's probably not
penetrating into this wall

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and it's probably not
bouncing off of this wall,

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it's probably constricted

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to be only in the plain of this wall,

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so there's gonna be no
horizontal acceleration.

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And if that doesn't make sense,

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it's because there's no motion

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in the horizontal
direction, left or right.

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There's no velocity, change,

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at all, in this horizontal direction,

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because the shoe's not gonna be moving

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in that horizontal direction,

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and it continues to not move
in that horizontal direction.

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So our acceleration horizontally

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is just zero equals the net force,

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divided by the mass.

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Alright, the net force in the x direction.

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What are we gonna have in the x direction?

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Well I've got fn pointing to the right,

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so again that's a positive force

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I'm gonna consider right
word to be positive,

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and I've got this F4, part
of it points to the left,

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so just like before, I've
gotta break this force up,

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I've gotta figure out how much

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of this force points horizontal,

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and how much of this
force points vertical,

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to get this F4 in the x direction,

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which is what I plug into
this formula up here,

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because I need this component here.

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This is the horizontal force of F4,

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not the vertical force.

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I don't plug the vertical
force in anymore,

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because this vertical force is
not part of the x direction,

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we're considering Newton's
Second Law for the x direction,

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so to solve for F4x, I'm
just gonna again use sine,

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because this angle, the
opposite of this angle is F4x.

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I'm gonna use sine of theta,
oh sorry, sine of phi.

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I'm gonna take sine of phi,

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that's gonna equal F4 and the x,

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divided by the total amount, F4,

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I get F4 in the x, is gonna
be F4 times sine of phi,

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and now I can use this up here,

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but you gotta be careful
with sines F4x points left,

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I'm gonna consider that a negative force.

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So if F4 sine theta
represents the magnitude,

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I'll write this as negative F4, sine, phi,

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sorry, I keep saying theta, I mean phi,

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I multiply both sides by m,

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I'll get zero again on the left hand side,

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equals, I've got Fn minus F4, sine phi,

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and now, when I solve this
for Fn, the normal force,

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I'll get the Fn, I'll add this
F4 sine phi to both sides,

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and I'll get that this normal force

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is gonna equal F4 sine phi.

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And that makes sense.

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It makes sense because,

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what these surfaces are doing,

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the reason why you're
getting a normal force,

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is these surfaces are exerting
whatever force they have to,

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to prevent any penetration
of this surface.

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So, if this F4x is pushing
in to the surface with F4x,

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right, if that's the force
we're pushing in with,

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Fn's just gotta equal that.

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It's gotta match that so that

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there's no acceleration horizontally.

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There were no other forces.

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We could, now you know
what to do if there were,

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if you wanted to step this up,

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you could add another force here,

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we'll call that F5, that'd
be another force this way,

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we'd have another F5, you
know how to handle that,

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now you'd come over to here,

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that's pointing to the
left, so you do minus F5,

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you'd come down here,
this would be a minus F5,

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you'd add that to both sides,

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that'd be a plus F5.

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What if we added a vertical force?

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What if we added another
vertical force this way

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to the shoe and we called that F6?

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Well that wouldn't impact
the normal force at all.

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This force F6 does not affect
how much these surfaces

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are getting pushed into each other.

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So I wouldn't include
that over here at all.

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That's a vertical force,

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it wouldn't affect the
normal force this time.

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Also note, gravity's not even affecting

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the normal force this time.

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'Cause gravity's exerting a
force in the vertical direction,

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and our normal force is in
the horizontal direction.

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So, long story short, normal
force is not always mg,

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the normal force will only
exist, it'll only be non zero

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when two surfaces are in contact
and pushing on each other.

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You can change what the normal force is

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by adding forces into
or out of the surface,

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exerted on that object.

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And if there's a force at an angle,

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when you're finding normal force,

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make sure you only use the component

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that's in the same direction
as the normal force,

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'cause that's the only
one that's gonna affect

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the normal force when you solve
using Newton's Second Law.