Leah here from leah4sci.com/MCAT and in this video we'll talk about Force Equilibrium and
Free Body Diagrams in preparation for the MCAT. Now I teach this topic with caution
because free body diagrams can be very time consuming and on the MCAT you don't have that
time. So instead I'll show you what to do and I want you to utilize this in practice
but when it comes to your practice exam and you real exam, I want you to try to simplify
this as much as possible so that you don't waste any time.
When we're dealing with equilibrium, you'll often see an object that isn't moving or does
not have a net force in any direction. You can call the subject the equilibrium to imply
that the forces in the different directions are all equal or they all cancel out. But
there are different types of equilibrium to consider. Translational Equilibrium has to
do with motion, so that means that if you have an object that has a force in one direction
and it won't be moving but then you have a force in the other direction canceling it
out that's considered Translational Equilibrium. But, that means the object doesn't have a
net force, it necessarily mean that the object is stationary.
Let's use a simple block where the demo force is gravity and the upward force is the normal
force. This block is not moving in the Y-direction because it is in equilibrium, meaning the
upward force and the downward force perfectly cancel out or simply the sum of the force
in the y-direction is equal to zero. But now for example, if the block is moving to the
right or the left with constant velocity that means the change in velocity is equal to zero
and change of velocity over time is equal to acceleration. So if delta v over t is zero
then f equals ma, a is zero, force is zero, this object is still in equilibrium but this
is considered dynamic equilibrium, it doesn't have a net force, it still has a constant
velocity and is moving. To say that an object is not moving at all, that's when you wanna
talk about Static Equilibrium where you have the constant velocity of zero, it's static,
it's not moving.
When setting up this equilibrium equations, it helps if you omit those signs. For example
the positives and negatives and instead, think of what the sign implies so you have a logical
diagram because the signs can confuse you but the logic behind the signs will help you
with your drawing. A standard free body diagram is basically taking your object out of the
story, out of the question and draw arrows to represent the different forces acting on
this object. I like to draw a free body diagram with either a square, a block or a dot and
this block can represent anything, from a person to a car to a hammock, any object is
easier to draw than a block, than trying to waste time drawing pictures especially if
you're not an artist, because this is a Science subject, not art.
Now regardless of the size of this object, unless you're specifically being tested on
central mass or central gravity, assume that the entire mass of the object is concentrated
in the center of the block or the dot that you're drawing. And that means all the forces
can come out of that one object instead of having to figure out where to draw it on a
complex item. See this block is resting on a table we'll show that the downward force
is the force of gravity or mass times the acceleration of gravity, the upward force
is the normal force because the normal force applies to any object resting on a surface.
Now let's see the wind is attempting to force to the right. We will draw an arrow to the
right and call this force of the wind and you can put whatever subscript you want under
the f as long you know what you're referring to. Some of these are common, Fn normal force,
Fg gravity but in the last video we were Fp, the force of a person, Fw the force of the
wall, as long as you know what you're talking about, you're fine. But now let's say the
block isn't moving despite the wind trying to push it, we can show that the force of
friction is countering the force of the wind. Now this object here is in equilibrium so
all you do is set this equal to each other regardless of positive or negative and let
me show you how it works out.
If I tell you that the block has no motion or acceleration on the X-direction, we can
assume that the sum of the forces in the x-direction is equal to zero and since the forces are
force of friction and when in the opposite direction we can say that the force of friction
equals force of the wind. This is without using any positives or negatives, just logically
saying if one is to the right and one is to the left they're opposite and therefore they
are equal to each other if there is no motion. If you want to get technical you can set it
up as sum of the forces in the X-direction equals zero which is equal to positive to
the right so that's gonna be force of the wind, negative for the left, force of friction
and when you solve it, you'll actually get the same equation we wrote earlier but jumping
straight to the setup that doesn't have any signs is faster if you understand how to set
this up. Now we can do the same thing for the Y-direction.
We know the block isn't moving so the sum of the forces in the Y-direction are equal
to zero, the actual force is force normal and I write Fn so it's not to confuse it with
the capital N for Newtons. The upward force is force normal, the downward force is force
gravity, they are equal and opposite to each other so we set the equal to each, that's
the fast way now let's show it the other way.
If the sum of the force is in the Y-direction is equal to zero, that is equal to the normal
force upward, positive minus the gravitational force downward so that's negative and ones
again if we re-write this equation we get what we already have. Now imagine we have
the same block sitting on a frozen lake that has a low amount of friction so that the force
of the wind still cancels the force of friction but there is a net velocity on the object.
So let's say there was a strong gust of wind and the object started moving and then friction
opposes it just enough to cancel out any acceleration but the block is still moving at a constant
velocity. The block is still in translational equilibrium in this case dynamic equilibrium.
We can still set up the free body diagram because the sum of force is equals zero even
though it's moving at a constant velocity.
Other free body diagrams that you may see is, let's say this block now represents a
toy wagon and you have a kit moving it at some angle feda. So that will be your angle
feda. If it's being pushed or pulled at an angle the way you set it up is going to be
the same, but if the force is now in a slightly upward direction, you're dealing with something
between the X and Y components and you want to break it up so you can clearly see what's
happening in the X and what's happening in the Y direction. Down still force in gravity,
up still the normal force. But now we have an additional component add in the y direction.
We take that force either a push or pull at an angle we have to break that up into the
force x and the force y using trigonometry and now when you set up the equation you have
the sum of the forces in the y-direction is going to be the force applied or the green
force positive, the normal force positive and the gravitational force negative and that's
still equal to zero because the block is still moving upward down. But now when you setup
the x-direction you can say the sum of the forces in the x-direction, let's call this
as applied X so F-a-x, is going to be positive F a x which you would set up as F applied
cosine of feda and then if you have friction, you'd set that up in the opposite direction.
If you don't have friction then you have F of x equals mass times acceleration.
The scary free body diagram for most students is the inclined plane and that's because we're
dealing with an object that is not sitting on the x or y axis rather somewhere between.
Now I break this down in detail in the Trigonometry video which you can find on my website at
leah4sci.com/MCATMath and that's where you see exactly why I'm about to do what I'll
do, now I'll just give you the overview. All we do is switch the axis so that X is at angle
but parallel to the inclined plane and Y is also an angle but perpendicular to the inclined
plane.
In the Trigonometry video I proved to you why given an angle feda X is going to use
Sin rather that Cosine and why we'll use Cosine rather than Sine but everything else is still
the same. Gravity does not change just because the incline changes and so gravity go straight
down. If gravity go straight down and it's between the X and Y component you have to
break it up so that your force of gravity in the X direction is going to be Sine of
your angle and force of gravity in the Y-direction is going to be Cosine of the angle. Now we
don't care force gravity in the Y-direction, we do care about the normal force which is
equal to mass times gravity in the y direction, that's where we care about the cosine component.
And if you have friction, that goes in the opposite direction to the X component.
Another confusing factor here is when to use positive and when to use negative. Don't worry
about the signs that you're given, turn every sign into something logical. So for example,
gravity is a negative acceleration because it goes downward. Don't worry about the negative.
Gravity in this case goes to the left. Friction is negative because it opposes the direction
of motion. But in this case friction is towards the right so if I try use negative it'll be
very confusing. Now if I tell you that this block is not sliding down the incline plane,
then I can assume that the sum of forces in the X direction are equal to zero. If the
sum of the forces in the X-direction equals to zero then I can say that force of gravity
in the X-direction is equal to the force of friction and I don't have to worry about which
one is negative or positive because again this is where it gets confusing. Instead,
in this case gravity goes to the left because the way the incline plane is drawn goes left.
Friction goes to the right so don't worry about your negatives and positives and just
set it up this way. If you do have to set up a more complex equation pick one and stick
with it, as long as you're consistent it'll work out. I can choose this to be negative
and this to be positive making the gravitational acceleration negative but the friction positive
and then you typically wanna choose positive y upward and negative y downward, ones again
setting those equal to each other.
Notice here that the normal force is mg with a component of gravity so that's mg cosine
of feda which perfectly balances the force of gravity downward setting the sum of the
forces y equal to zero. Again for the breakdown of why will you sign it cosine opposite to
the normal convention, be sure to see the Trigonometry video where I show it in detail.
Don't take that explanation as something you have to apply each time but it needs to be
intuitive so that you don't waste time being confused on this.
Let's do one more free body diagram. Imagine we have a hammock suspended from two trees,
that'll be the hammock and we a student here laying in the shape studying for his MCAT.
If you want to calculate the sum of forces here, the first thing you want to do is trend
this entire system into one point. If you take the time to draw the hammock on the MCAT
you're wasting too much time. Here you have tree 1, tree 2, here we have the student with
a hammock as a dot and here we have the ropes that support the hammock. Let's make this
bigger so that we can mark it up. Ones again we set about with these sum of the forces.
Downward we have force of gravity the student in the hammock. Upward in this case, we have
not the normal force but the force of tension in the ropes that support the hammock. We
have two ropes sitting at some angle between the x and y component and so we have to break
them up into their vectors.
Assuming the ropes are at the same angle, the sum of the forces in the X direction equals
zero, coz the hammock isn't moving. That means the tension to the x of one rope equals and
balances the tension x in the other rope. We can simply call as positive x and negative
x. Assume they cancel out and don't worry about the math. But the tension in the y direction
can get tricky. The sum of the forces in the y direction equals zero and you can set it
up as force of gravity is equal to two time tension in the y direction. The reason it's
two is because half of it goes to one rope and half of it goes to the other rope and
obviously you'll find this using the angles given in Trigonometry but don't worry we'll
tackle tension by itself in its own video.
You can find my entire series on MCAT Forces along with cheat sheet and practice quiz by
visiting my website leah4sci.com/forces.
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