– 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.

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

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

Great Video!

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

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.

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

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

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

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

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!

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

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

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.

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

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

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

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.

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