Leah here from leah4sci.com/MCAT and in this video we'll talk about Normal Force as it
shows off in MCAT Physics.
Normal Force is one the less intuitive forces to learn because we take it for granted and
don't always realize it's there.
Say you have a block resting on a table, we know that the downward force of the object
is the weight of the block or the force of gravity.
But if the force of gravity is equal to mass time the acceleration of gravity you'll expect
the block to accelerate through the table through the floor and straight down to the
center of the earth which is the source of that gravitational attraction.
So why doesn't everything just sink through the surface it's resting on as where the normal
force comes in.
I like to think of the normal force as something keeps things normal because it's not normal
for stuff to just go through the table.
The normal force is a force that acts perpendicular to a surface that an object is resting on.
So when a box sits on a horizontal table, the perpendicular force is upward which is
exactly opposite of gravity.
That puts the box in equilibrium in the Y-direction setting the sum of the force as Y is equals
to zero.
And this is how most students have it memorized where force normal or simply N for the normal
force is equal to mass times gravity.
The equation looks true in this case but not in every case so be careful, don't memorize
normal force as equal to mg, understand that the normal force is simply perpendicular to
the surface that it rests on.
For example, say you have the same table, but now I'm sitting under the table and I
am pushing the block up and holding the block against the table, in this case I can say
F of p or the force pushed is what I have upward and the force in gravity is still acting
downward.
If my force upward is the same as my force downward, then all I have in the Y-direction
is the sum of the forces in the Y-direction equals F of p upward minus F of g downward
and I have nothing else.
But now let's say I'm pushing so hard that my force upward is greater than force of gravity.
For example if the block is 10 Newtons and I force with the force of 15 newtons, where
is the other 5 newtons is gonna go?
Well, in this case I wanna set it up as follows.
We have the force of gravity downward, the force of pushing upward and the difference
is the normal force which is perpendicular to the surface it's resting on.
If the block is resting upward on the bottom of the table, then the normal force should
also be downward and this is how we'll set it up.
The force push is equal to force of gravity plus the normal force.
I said the upward push is 15 newtons and gravity is only 10 newtons so we're assuming that
the object has a mass of 1 kilogram times 10 meters per second squared for the acceleration
of gravity gives me 10 newtons plus whatever I have for the normal force.
So we'll call that plus X.
If we subtract 10 newtons from both sides then we get a net of 5 newtons which is equal
to the normal force and this is why I write F or n so we don't mixed it up with the N
for newtons.
In this case the normal force is not equal to the force of gravity because of the way
it's setup.Well this isn't as likely to come up, what you'll more likely to see is an incline
plane with an object resting on the surface.
In this case we'll have some angle feda.
Remember that on an incline plane we change the X and Y coordinate system to have X parallel
to the plane and Y perpendicular to the plane.
If we setup a free body diagram, then the downward force is the force of gravity which
is in between the X and Y components, so we'll break it up where the force of gravity in
the X direction is force of gravity sine feda, if you're not comfortable of why we chose
sine, go back to my Trigonometry video on my website at leah4sci.com/MCATMath and then
take a look at the incline planes video in the Kinematics series leah4sci.com/Physics.
So we have the gravity broken down into force of gravity in the X-direction using sine and
then force of gravity in the Y direction using the cosine of feda.The reason this is important
is we look at the way the object is sitting on the incline plane, the normal force has
to be perpendicular to the surface which is exactly equal on the force of gravity in the
Y-direction.
So here's how we'd set it up; the sum of the forces in the Y-direction equals zero because
if the object, it'll only move in the X-direction up or down the plane, but the object is not
being physically lifted off the plane or going through the plane and that means we have the
normal force on the upward direction minus the force of gravity in the Y-direction equals
zero or simply Fn equals negative Fgy.
And so solve for the normal force in this case, we have the normal force is equal to
the force of gravity or mass times the acceleration of gravity times cosine of the angle feda,
it's not just mg it's mg cos feda.
This is critically important when you're dealing with friction which we'll tackle on a separate
video but for the force of friction, you have mew, sub k or sub s times n which is the normal
force and you wanna make sure you don't just use mg because that will throw off your calculations,
you wanna make sure you remember to use cosine of feda to make sure you're taking that full
y-component of gravity on the plane.
Be sure to join me in the next video where we talk about Gravitational Force near the
surface on the earth and objects orbiting or other planets.
You can find my entire series along with my Forces series' cheat sheet and practice quiz
by visiting on my website, leah4sci.com/Forces.
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