Say you want to body slam an outdoor shed and it doesn't move. But you do the same thing
to your niece's doll house and you send the teddy bears flying. In this Leah4sci MCAT
video, we'll look at the Forces of Static and Kinetic Friction. Friction is a force
that opposes motion and we have two types. Static Friction opposes the motion of something
static or stationary, meaning it prevents a non-moving object from starting to move.
Kinetic implies motion, therefore kinetic friction will slow down an object that is
already in motion. Say you have a shed in your yard and it's just sitting there minding
its own business not exerting any forces, right? wrong! Don't forget that the shed is
on earth and therefore it's impacted by a downward force of gravity which is opposed
by an upward normal force as I discuss in the previous videos linked below at leah4sci.com/forces,
and there are no forces in the X direction until you come along and you push the shed
with your finger. Is the shed going to move? Not at all! But wait a minute! You just applied
a force in the positive X direction, that means there has to be a counter force in the
negative x direction to keep the shed stationary and this, this is what static friction is
all about. If you push on an object that isn't moving, that object is going to resist with
static friction with the exact amount of friction that you applied to the object so that it
doesn't move. Now say you apply even more force and body-slam the shed, you applied
a much greater force in the positive x-direction, but the shed countered with an exact equal
force in the negative X-direction, once again, it's not moving. Until a monster truck comes
along, slams into your shed and now the shed goes flying. Why does the shed move now? because
the truck is applying such a big force in the positive x-direction that no matter how
hard the shed tries, it cannot exert a big enough force in the negative x-direction,
and when we set up the free body diagram so that the sum of the forces are equal to force
of the truck minus the force of the shed, we do not get zero, we have more force in
the positive direction, static friction is overcome and the shed goes flying. Let's look
at this from an equation perspective. Static friction written as lowercase f sub s is the
amount of opposing force that an object can apply. The opposing force will always be less
than or equal to the force applied. If you hit the shed with a tiny force like your pinky,
the shed will counter with a tiny force of friction equal to your pinky. If you body
slam the shed, the opposing force goes up equal to the force of your body slam. But
if the opposing force is so strong that the shed cannot match it, that's when force of
static friction is less than the force applied, we overcome the static friction and the object
starts moving. If the force of static friction is equal to the force applied, then the object
is not moving. If the force of static friction is less than the force applied, the static
friction is overcome and the object goes into motion and that is because every object has
a maximum resistance. For example, a doll house will have a tiny static friction, a
shed will have a massive static friction. Why do we say less than or equal to? Why doesn't
the shed automatically apply a maximum static friction? Very simple! If you push the shed
with a pinky amount of force and the shed opposes with a body slam counter force, wouldn’t
the shed push you? What if the object is already moving? That's where Kinetic Friction comes
in which is lowercase f sub k to resist a moving object and ultimately slow it down
until it comes to a complete stop.
I like to go walking on my breaks and the recent blizzard didn't stop me. Problem was,
what came down as snow, piled up on the ground, melted and then froze into ice which means
I had to be very careful because if I stepped on the ice, I found myself sliding forward.
What happened? With every step, I had forward momentum and that means I had some force in
the positive x-direction. Say, instead of stepping onto a patch of ice on the sidewalk
I went ice skating. With my shoes instead of ice skates. If initially I found myself
sliding, assuming I didn't fall flat on my back, eventually the slide would get slower
and slower until I came to a complete stop. So what happened here? When I stepped onto
the ice and started sliding, I had motion in the forward direction and that means I
had some force applied, the force of me stepping onto the ice. However, the moment I stepped
onto the ice, the kinetic friction of the ice starts opposing my motion and if I'm sliding
forward, the ice pushes back, I slide forward a little more, the ice keeps pushing back
until the ice has applied so much kinetic friction that I ran out of energy, I ran out
of motion and came to a complete stop. Why doesn't this happen on concrete, or grass?
Very simple! Every object will have a coefficient of friction which tells you how hard it will
work to resist something moving on that object. That is represented by the letter μ. We have
μ sub s for the coefficient of static friction, how much something will resist an object starting
to move and μ sub k for the coefficient of kinetic friction, how strongly the object
will oppose something that is already moving. As you can imagine, ice will have a lower
coefficient of friction while something like wood would have a much higher coefficient
of friction. μ sub s will always be greater than μ sub k because it's much more difficult
for an object to start moving but once it's going, it wants to continue going. Think about
studying. It's hard to sit down but once you're already in the zone, it's very easy to keep
going. Objects are the same way. Which brings us to the equation for friction where the
force of friction, static or kinetic is equal to μ sub k or sub s times n. f is the force
of friction, k represents kinetic but an s would represent static is equal to μ the
coefficient of friction, sub k for kinetic sub s for static, times n which is the normal
force. Let's take a look at the coefficient of friction. If we look at waxed wood on wet
snow, μ sub s is equal to 0.14, μ sub k is equal to 0.1. What does this tell us? That
waxed wood like a sled to go sledding has 0.14 as its coefficient of static friction
but 0.1 as the coefficient of kinetic friction. Notice that static is greater than kinetic,
that is because it's much more difficult to get an immobile object to start moving than
to slow down an object that already has momentum. We expect something to slide easily on snow
which is why the coefficients are very low. But if I look at something like steel on steel,
the coefficient of static friction μ sub s is equal to 0.74 but μ sub k is equal to
0.57. Notice that I didn’t give you any units for the coefficients, and that's because
it's a ratio of how much it wants to move versus how much it doesn't want to move and
the ratio doesn't have any units. Luckily synovial joints have a μ sub s of 0.01 and
μ sub k of 0.003 because we need to be able to move. Back to our equation, μ sub k times
n, remember that n is the normal force, F sub n or simply n which is equal to mass times
the acceleration of gravity. To set up a problem dealing with friction, we start with our free
body diagram or just simply put a point and apply all your forces. In the downward direction,
we have the force of gravity, in the upward direction we have the normal force. In the
positive direction we have the force applied, this would be your motion or your attempt
at motion, body slamming a shed. In the negative x-direction we have the force of friction
opposing the force of motion. When dealing with an object on a horizontal plane the y-direction
is very simple. If the object is resting on a surface meaning it's not going down into
the ground or up into the air, then the sum of the forces in the y-direction are equal
to zero and that is because the force of gravity is equal to the normal force meaning mass
time the acceleration of gravity is equal to n. The problems typically come up when
we look at forces in the x-direction.
For static friction, if an object is not moving and the sum of the forces in the x-direction
are equal to zero and that is equal to the force applied minus the force of static friction
which you can simply write as force applied is equal to the force of static friction and
force applied is equal to μ sub s times n. In the case of kinetic friction, in the x-direction
when the sum of the forces is not equal to zero, that's where you get your mass times
acceleration recognizing that acceleration is going to be negative because it's technically
a deceleration. If I'm sliding on the ice and the ice is slowing me down, my acceleration
goes lower and lower and lower until I come to a complete stop.
Ready to try a practice problem? If YES, first give this video a thumbs up, subscribe to
my channel and make sure you hit that bell notification so that you don't miss out on
any new videos and then let's take a look at this.
A hockey puck is sliding on the ice at 6 meters per second, how far will it travel before
coming to a complete stop. We're given two variables, mu sub k is equal to 0.17 and mass
of the puck is equal to 160 grams. Before you even start the problem, ask yourself,
what are we looking for? How far will it travel tells me I’m looking for a delta x and coming
to a complete stop tells me that v final is equal to zero knowing that v initial is 6
meters per second. Let's set up a free body diagram starting with our hockey puck and
forces in the y-direction. In the downward direction we have the force of gravity. In
the upward direction we have the normal force. They cancel out so we know that the sum of
the forces in the y-direction is equal to zero, the puck isn't going down into the ice
or up into the air. For the puck to come to a complete stop, something has to slow it
down and that means if we have a velocity initial of 6 meters per second, velocity final
of zero meters per second, there has to be an acceleration because the change in velocity
over the change in time is equal to our acceleration, in this case negative acceleration because
it's going down. What causes that deceleration? That is the force of friction and that is
what we're solving for. The force of kinetic friction is equal to mu sub k times n which
is also equal to our mass times acceleration. We're about to bring in Kinematics and if
that's something you're not comfortable with, go back to my other video series at leah4sci.com/kinematics
linked below. How do we know that it's equal to ma and not zero? simple! we have a change
in acceleration. How does this help us solve for delta x? Very simple, if we find our deceleration,
we can use that to find our delta x which makes it a two-part question. Let's start
by solving for and therefore isolating acceleration. This means we bring m over to the other side
giving us a new equation, acceleration is equal to mu k n which I'll remind you is mu
k times mg divided by m. That was convenient, m drops right out, giving me mu k times g.
We were given mu k of 0.17 and we know that gravity is equal to 10 meters per second squared.
Using the decimal trick that I teach in the math videos linked below, we get our acceleration
is equal to negative 1.7 meters per second squared. Don’t forget the negative because
it's slowing down. Now that we have acceleration, we turn this into a new problem where our
given is as follows; acceleration is equal to negative 1.7 meters per second squared,
v final is equal to zero, v initial is 6 meters per second and our unknown that we have to
solve for is delta x. Using the which kinematic equation trick that I teach in the video linked
below, we're going to use the time independent equation as follows: V final squared is equal
to v initial squared plus 2a delta x, so let's isolate for delta x. Subtract v initial squared
on both sides giving us v final squared minus v initial squared is equal to 2a delta x.
Remember that v final is zero and since a is negative, instead of dealing with that
negative sign, just say this is negative, this is negative, they cancel out to give
us a positive delta x. Let's keep isolating, divide both sides by 2a which tells us that
delta x is equal to v initial squared over 2a. Let's plug in the numbers, v initial is
6 meters per second, square the whole thing, divided by 2 times the now positive acceleration
1.7 meters per seconds square. Let's ignore the units for a moment. 6 squared is 36, divided
by 2 times 1.7, this is where you have to get a little creative, you can either recognize
that 2 times 1 is 2, 2 times 0.7 is 1.4 giving me 3.4 or you can round it and say, 1.7 is
close to 2 and 2 times 2 is 4 so something under 4, but I'll use my first trick for 3.4.
36 over 3.4 is complicated but remember on the MCAT close enough is good enough which
means if I estimate this to be about 36 over 3.6, you can either recognize that 36 is 10
times larger that 3.6 or combine two of the tricks that I teach in the math series linked
below as follows; move the decimal one space to give you 360 over 36, cancel out the 36
to give you 10, cancel out the 36 to give you 1, 10 over 1 is 10 giving me a delta x
is equal to 10, we expect the units to be meters but let's take a look. Meters per second
squared means we square the whole thing giving me meters times meters over seconds times
seconds. If we're dividing by meters per second squared, the seconds squared cancels this,
meters cancels the green meters, leaving me just meters as expected.
While this problem had a lot of math, I wanna argue that it was pretty straightforward.
Hopefully you’ll agree with me after going through it one or two more times. But what
happens if you're given a problem on an inclined plane where motion is going down the plane,
friction up the plane but gravity is going straight down which means gravity is also
opposing friction. That's exactly what we'll cover in the next video which you can find
linked below or by visiting my website leah4sci.com/forces . The link again, leah4sci.com/forces.