- Good morning, Bo could
you please read the problem
and Bobby could you please translate?
♫ Flipping Physics ♫
- A car with anti-lock
brakes driving on snow has
an initial velocity of
8.9 meters per second
and slows to a stop in 3.12 seconds.
- Velocity initial equals
8.9 meters per second
and change in time is 3.12 seconds.
- Also, the car stops so
the final velocity is zero.
- Yeah.
- Bo could you please continue?
- Determine the coefficient of friction
between the tires and the snow.
- Mu static equals question mark.
- Hold up, shouldn't
it be kinetic friciton
because the car is moving?
- The car has anti-lock
brakes so it's static friction.
Didn't you watch my video project?
- Yeah, right, yep.
- Okay, now let's see what this
problem actually looks like.
(banjo music)
Billy, how shall we begin?
- Actually, I don't know where to start.
- [Mr. P.] Okay, so well then
what are we trying to find?
- We're trying to find the
coefficient of static friction
between the tires and the snow.
- And what is an equation we have that has
the coefficient of static friction in it?
- The force of static
friction maximum equals the
coefficient of static friction
times the force normal.
- Therefore, we are dealing with forces,
which means class, we need to draw a
- [All] Free body diagram.
- Billy, what are the forces
in the free body diagram?
- The force applied is to the right,
the force normal is up, the
force of gravity is down,
and the force of static
friction is to the left.
- Class, which one of these forces is
not actually acting on the car?
- The force applied.
- Yeah, just because the
car is moving to the right,
doesn't mean there is a force
pushing it in that direction.
- I forgot, it is the inertia of the car,
which tries to keep the
car moving to the right.
- Absolutely correct, remind me Bobby,
what are the three things you need to
remember about the direction
of the force of friction?
- The force of friction is always parallel to the surfaces.
It opposes sliding or motion, and is
independent of the direction of the force applied.
- You can see the force of
friction is parallel to the
ground and because the
car is moving to the right
and the force of static
friction is to the left,
you can see that the force
of static friction opposes
motion, or opposes sliding,
and it is independent
of the direction of the force applied;
you can see that because there is
no force applied in the free body diagram.
Bo, we have drawn our free body diagram,
what do we do next?
- Well, we don't need to break
any forces into components
because they are all directly
in the x or y direction,
therefore we don't need to
redraw the free body diagram
which is nice, therefore
let's sum the forces.
- Okay Bo, please pick a
direction and sum the forces.
- Okay, let's do the net
force in the y direction
equals the force normal
minus the force of gravity
which equals mass times
the acceleration of the car
in the y direction, which
is zero because the car
isn't moving in the y direction,
therefore the force normal equals the force of gravity
and the equation for the force of gravity is the mass
of the car times the acceleration due to gravity.
- But we don't know the mass of the car.
- Ah, I can feel it's almost party time.
(laughter)
- So, we put the equation
for the force normal
in our equation holster,
and I want to pause
for a moment and identify this;
we summed the forces in the y direction
and we got the force
normal is equal to the mass
of the object times the
acceleration due to gravity,
and we have actually done this exact thing
many times now, right class?
- [All] Yes.
- As students, ya'll tend to
look for repeatable patterns,
and many of you therefore
are probably assuming
that this is always true,
but this is not always true.
The force normal will
not always equal the mass
of the object times the
acceleration due to gravity.
There is no equation for the force normal;
in order to find the force normal,
you need to draw a free body diagram
and sum the forces, every time.
Back to the problem, Billy could you
please tell us what to do next?
- Well, the net force in
the x direction equals
the negative of the
force of static friction;
negative because it's to the left,
and that is the only
force in the x direction,
so that equals mass times acceleration.
We know it is the maximum
force of static friction
because I showed in my video project that
the anti-lock braking system, or ABS,
tries to maximize the
force of static friction,
so we can substitute in the coefficient of
static friction times the force normal
for the force of static friction
and from our equation holster we can
substitute in the mass
of the vehicle times
the acceleration due to gravity
for the force normal and--
- Everyone brought mass to the party.
♫ Everybody brought mass
- Yes, we can be
equitable, we can take mass
from everybody, therefore
notice that the mass
of the car is irrelevant
as far as determining the
coefficient of static
friction is concerned.
Billy, please continue.
- After the mass party removes
mass from the equation,
we are left with negative
coefficient of static friction
times the acceleration
due to gravity equals
the acceleration of the
vehicle in the x direction.
Dividing both sides by the acceleration due to gravity
and multiplying the whole equation by negative one,
gives us the coefficient
of static friction equals
the negative of the
acceleration in the x direction
divided by the acceleration due to gravity.
But, we don't know the
acceleration in the x direction.
- True, we don't know the
acceleration in the x direction.
Yes, Bobby?
- However, we know the
acceleration in the x direction
is constant because the force
of static friction is constant
therefore, the net force in
the x direction is constant.
- So, we can use the
uniformly accelerated motion
equation, velocity final
equals velocity initial
plus the acceleration
times the change in time,
subtract velocity initial from
both sides of the equation
and then divide by
change in time to give us
acceleration equals velocity final minus
velocity initial divided
by change in time.
- Which is the equation for acceleration.
- Yeah, that's right, thanks.
- You're welcome.
- Oh, we can know substitute
in numbers to solve
for the acceleration, right Mr. P.?
- Sure, Bo, Bobby could you
please substitute in values now?
- So acceleration equals...
Zero minus 8.9 divided by 3.12
which works out to be,
2.8, I'm sorry,
negative 2.85256
meters per second squared.
- Billy, please finish.
- Okay, let's see then, the
static coefficient of friction
equals the negative of negative 2.85256
divided by 9.81,
which is 0.290781,
or, with sig figs, 0.29,
what are the dimensions
for the coefficient
of static friction?
- it doesn't have any,
jinx, you owe me a soda.
- Thanks?
- Right, remember the
coefficients of friction
do not have dimensions,
now I actually preformed
nine different trials of this experiment
and each trial had a slightly
different change in time.
This meant I also determined
slightly different coefficients
of static friction for each trial.
The range of values for the
coefficients of friction
was between 0.28 and 0.32 and the average
value worked out to be 0.30, that means,
for this specific situation,
0.30 is the average
coefficient of static
friction between these tires
and a snow covered road.
Thank you very much for
learning with me today,
I enjoyed learning with you.
