- 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.