– 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.

## Only registered users can comment.

1. TheHammer says:

Is it funny that I just wrote a test couple hours ago with a very similar question? Loving your videos!

2. Frank Thomas says:

First time I've noticed that the socks are symmetrical.

3. Eleazar Almazan says:

Great Video!

4. Dave Felix says:

Awesome videos man! Really helping me out! You make it a whole lot easier

5. JusaStronomer says:

Was Billy's video project about Anti-Lock Brakes actually produced and uploaded to YouTube? It's not in the Newtonian Physics playlist and I couldn't find it anywhere with a quick search.

6. andres villa says:

i do not understand why you still have just 9k subs .u must be like running on like a 5million . no props on oct 2 2016 you would have that

7. Mateo Agudelo Toro says:

How come this guy isn't a superstar on YT?

8. Jinu Seol says:

AWESSSOME!!!!!!!!! 🙂 Always thank you!

9. mohamed bensalah says:

Love ur videos man,, keep it up,, thank u

10. john s says:

I'm a war veteran and engineer turned physics teacher. I teach in rural thailand but will be in bangkok soon. Your videos are well thought out and creative. I wish I could donate. I've been working in these poor public schools for too long. >1k per month!

11. RickTrajan says:

ey I need to watch all your videos to understand all these. but I think it would be well worth it.

12. ibrahim ismail says:

Thank you so much great work please continue, you are superstar…

13. Lucas M says:

How isn't there an applied force Fa? Friction is pushing to the left. How can the car go to the right if there's nothing pushing it to the right? Maybe the motor is Fa.

14. windthorpe9 says:

5:31 hahaha "Everyone brought mass to the party"

15. Dragonpops says:

Great video! for interest, what type of tires did you have for this test? Winter tyres or all seasons (M&S)?

16. Science Seeker says:

Please give me an answer why did we say there is no force applied .. i want to show to my teacher .. please type the full answer here and thanks in advanced

17. atcobra65 says:

The video is totally worthless. You don't live in the Midwest. All that bullshit goes out the F'ing window when an 18 wheeler passes you at 70 mph and your at destination F"d". Tire static coefficient, anti lock brakes, tread depth , driver reaction time. Hmm.. is there hard pack ice under the snow, what is the frost depth, is there hard pack snow. What kind of tires are on the vehicle, brakes been replaced? Is the vehicle a Ford 150 , Toyota Camry . Guessing I'm too f'ing dumb too ask those kind of questions. Drive fast, take, chances.

18. Sahilbhai Rahish says:

Please explain in detail ,your way to teaching is awesome ,i love to watch your video series