umm, i just used the pretty standard intuition but his limit thing made me clear the concept of limit now 🤣.

How can you see the dislikes?

@@jenniferratto9232 cuz it was 3 years ago

14 ??

because obviously an engineer would skip to the end of the video and use random numbers on the screen without noticing that it' s the frictionless case....

Occupational Hazard

I'm a physics undergrad, not a teacher. And still my friend who's an artist says I'm better than her at drawing straight lines.

Excellent! I wish you well in your future endeavors. Cheers, Dr. A

i am a middle school student.....

@@reemabansal9594 what does this comment have to do with anything?

"One should make things as simple as possible, but not simpler." - Albert Einstein Cheers, Dr. A

Thanks for the kind words. Not writing backwards, of course. Answer here: www.learning.glass Cheers, Dr. A

Kerry-Anne, Thanks, that's great to hear. Good luck with your coursework. And remember to take some time to investigate "other" classes besides your major. College is a very unique time in your life where you can explore some very different subjects. One of the most interesting classes I ever took in college was called, "Experimental poetry." Now that was different. And super awesome. Cheers, Dr. A

Thanks Professor for the advice - I was thinking of doing Latin or even German as these are very interesting topics/areas that I know I would enjoy. I am also very fond of Art history so I will definately consider these options.

Krzysztof, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

Now you know. Cheers, Dr. A

he is THE TIIIIIIME WIZARD

maybe he is not writing backward but flipping the video horizontally

@@linhly5967 but there is student in there

@@abdullahh.mesawa5206 he flipped it, you can tell by the apple logo on his computer :0

@@abdullahh.mesawa5206 It's called the Learning Glass, he writes behind the glass and then there's a projector showing the flipped video real time to the students

You don't necessarily know that. It is simply a special case of interest to this problem, where the car drives on the banked curve at a specific speed, where the traction force is zero. The horizontal component of the normal force is enough to push the car inward to make the curve. If there is a traction force, it could either act up along the incline, or down along the incline. If you take the curve "too slow", it acts up along the incline. If you take the curve "too fast", it acts up along the incline..

If you take the banked turn at the critical speed, and you already started, when you let go of your steering wheel, your wheels will align with you continuing on the curve. Rather than finding the straight position as they ordinarily do, the wheels will find the position to keep you on the banked turn. You'll notice that it takes considerably less effort to hold the steering wheel in this position than it does when you take a curve on a flat roadway. If you rigidly maintain the neutral straight position with your steering wheel, your car will climb the banked curve, and will use a traction force to do so.

Excellent. Just in time. Cheers, Dr. A

Definitely not sound proof. Cheers, Dr. A

Thanks! Love the awesome sauce comment. Cheers, Dr. A

i didnt take physics classes in high school.

This video presents a classic physics example. The goal is to design the curve such that zero friction is necessary to make the turn. But your question is different and quite interesting. Let me restate it: Pretend you take a left turn around a banked curve at high speed. Are you steering left, straight, or right? I don't know the answer to this question, but it would clearly depend on several factors, including what your rear wheels are doing. If they are staying in-line, then I would say you need to steer left (the inner radius of the loop is slightly smaller than the outer radius). If the rear wheels are skidding out like a dirt track racer then you would need to steer right. Something in between the two, and you might need to steer perfectly straight. Good question. Cheers, Dr. A

Thank you for the response, looking over my question would need several factors in order to be answered. What I believe just by picturing it within my head is that it is harder for the rear wheels to lose traction going around a banked turn, as the outer rear wheel (which is most prone to spinning) is being pushed harder onto the surface than the inner wheel. That's just in my head though, sorry for probably giving you a headache with my question...

Don´t take under-/over-steer into consideration, at normal speeds (and with normal I mean relatively low lateral dynamic, like below 5 m/s2) that effect is negligible compared to the other forces at play. If you drive on a banked left corner at the neutral speed you have no lateral acceleration: in this case, if you take your hands off the steering wheel you will see it setting itself at the right angle for the car to stay in the lane, just like it recenters on a flat, smooth and straight road (actually even better, as the car can´t drift away: fun fact, when driving like this you can manage your height in the banking simply by changing your speed by a few km/h). Anything faster than the neutral speed, and you will have to apply increasingly more left (inward) steering to counteract the higher lateral acceleration, just like you would do on a normal flat left corner (apart from some minor non-linear effects that you need to compensate and tend to freak you out the first times). Anything slower than the neutral speed, and you will build up inward lateral acceleation due to the downward pull not being compensated with enough "centrifugal" force: that means that you will need to hold your steering and eventually steer right (outward) to effectively have the car "climb" up the banking wall; it might look like a countersteer (left corner/right steering), but this has nothing to do with oversteer as the lateral acceleration is really directed in the other way.

When does it states that Fn is equal to the y component of Fg? I guess that´s the root of misunderstanding, it isn´t ;)

Allison Benjamin, It would depend on your speed. If you go very fast, static friction is acting down the ramp. If you go very slow, it's acting up the ramp. If your speed is just right, static friction is neither up nor down (it's zero). Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

As the angle changes N changes, not mg

Thanks for reaching out. Black ice on the slot car set is not usually a problem (except for maybe a short circuit). Cheers, Dr. A

Very similar. With the mass on the inclined plane, we rotate our coordinate system such that x is along the incline and y is normal to it. That means we need to break mg into new x and y components. Here, we keep the coordinate system level, with x horizontal (which is along the radius of the car's circle). That means we need to break up the normal force into components. Cheers, Dr. A

@@yoprofmatt Thank you so much!, i get it now :)

Not only that, but how can a component of the normal force, which is a component of gravity in this case, be equal to the force of gravity itself? That seems to be counterintuitive.

@@vedantjhawar7553 You select whichever coordinate system is most convenient for you to solve the problem. In this particular solution, he selected a coordinate system with a y-axis that is parallel to gravity. This means the normal force has to be decomposed into its components, but gravity only acts in the y-direction. You could also opt to solve this problem in the coordinate system that is rectilinear with the road surface. But you most likely wouldn't want to do this, because then you'd also have to componentize the acceleration vector.

Samer Qansu You can decide what you want the x and y planes to be at the beginning of the problem, in this case he set the y plane to be the same direction as the mg, which means that the x component would be 0 since it is only going in the y direction, and they y component would be mg because it is going entirely in the y direction. If he set the y plane in the same direction as the normal force, and the x plane along the ramp, then he would have calculated the x and y components of mg because it would then be on a diagonal.

🚖 r1 and r2

Is is USUALLY the case, but not always. It depends on where your angle is defined. Always go back to SOHCAHTOA. Cheers, Dr. A

@@yoprofmatt thanks for the reply

@@ramk1985 Here's a way to remember it: "Cosine" = close (as in opposite of far). "Sine" = sign post. Raise the sign post up to the vertical where a sign post should be, and the sine of the angle is its maximum. When the angle is small, cosine gets you the component close to vector in question. Sine gets you the component projected onto the vertical sign post.

chrisvevo What is AP physics? Is AP similar to A levels in the UK? In the UK, once we do our GCSE's which is the eqivalent of SAT's in America. We either go to a college or study A levels which are required to go to university. A levels are very difficult as they require u to do 3 advanced subjects. Is that how AP classes work?

@@RAAAHUM AP stands for Advanced Placement in America. It's supposed to be a college level class. At the end of the year, there's a test, and if you do well on it some colleges will give you the credit for free if you go there. Basically a way to get a high school credit and a college credit at the same time.

@@julianreinhart504 Okay thank's BUT do u have to do 3 subjects to get into a college/university? i.e ap physics, ap maths and ap computer science for example. In the UK when studying A levels, you have to study at least 3 advanced subjects in order to get into a university. Also A level exams are as difficult as college/university exams, so yeah AP is basically another term for A level. Thanks for helping me understand.

@@RAAAHUM Happy to help out! But no, they are entirely optional. Most universities want to see some sort of advanced classes, but they aren't explicitly required.

@@julianreinhart504 Yeah A levels are optional to. Universities in the UK also accept something known as a level 3 BTEC qualification which are somehow equivalent to A levels even though no exams are required and it's 100% coursework which is an absolute joke.

When you introduce a friction force, you end up needing to solve a system of equations to determine your unknown normal force and friction force, that cause the car to follow a banked curve in the road. When we consider the friction force F to be zero, we are looking for the critical driving speed where we aren't depending on traction to cause the car to follow the curve. Friction will be present, but there is a critical speed where it is equal to zero, and you can solve for that speed with only one equation To work with friction, you end up solving the following two equations together for N and F. This assumes F acts inward on the banked turn, which means it also acts downward. If you get a negative number for F, if means you are driving less than the critical speed, and F really acts upward and outward, instead of downward and inward. N*sin(theta) + F*cos(theta) = m*v^2/r N*cos(theta) - F*sin(theta) = m*g Since the limits of F depend on N, you can then use it to calculate the maximum and minimum (if applicable) speeds at which you could drive on this curve. Max magnitude of F = mu_s*N.

Common misconception. Even though the tires are rolling, the rubber on the bottom of the tire is not skidding, thus we use static friction. Note: If you start to skid, then it becomes kinetic friction. Cheers, Dr. A

Got it, thanks!

@@yoprofmatt Thank you so much:)

Nope. It's only equal to mg on a level surface with no acceleration. Cheers, Dr. A

মঈনুল ইসলাম আবীর no not always

In ojbect on equal plan N=mg

Came here for typical calculation and advanced reasoning for friction

Check it out here: www.learning.glass Cheers, Dr. A

Awesome, thanks! btw, you saved me on my Physics test. Thanks a bunch!

Excellent! Hope you got an A. Cheers, Dr. A

Check it out here: www.learning.glass Cheers, Dr. A

Imad Shawish, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

You still do have to steer, and you do have to hold your wheel in place, because the car will generally cause the steering to default to straight, and then your wheels will introduce a friction force that causes you to no longer follow the curve. At the critical speed on a banked road, the only force inside the reference frame of the vehicle that you will feel, is a downward force that is perpendicular to the road. The net traction among the four wheels will add up to zero. It doesn't necessarily mean it will be zero in all four wheels, just that it will add up to zero. If the road is icy and has less friction, it is safe to drive at the safe driving speed on such a road.

Minh Hoàng Nguyễn, Nope, first time I've heard that one. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

I hope nobody understands you and Matt except Honda and Red Bull! Zandvoort is full of banks. F1 2020 勝つために ! からの挨拶 オランダ

Darth Negative Hunter, I hope not. I always try to teach my students that trying to do physics by memorization is a bad idea. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

Robert Overbeeke, Red Bull? Like the beverage. I'd love to help. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A

Harvey Rayner, Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass Cheers, Dr. A

@@yoprofmatt yh i think I will, thankyou

gattopazzo80, That does seem steep. I think our speeds are a bit high for the radius. I looked up offramp design, and typically for 25 mph you'll have a radius of 50 m or so. That would give us an angle of about 15º which seems more reasonable. Thanks for the tip. You might also like my new website: www.universityphysics.education Cheers, Dr. A

@@yoprofmatt it still looks a bit too steep as tang(15) = 27% cross slope; 10% is usually considered a dangerous longitudinal slope, and it doesn't impact lateral dynamics. Although some banking is definitively built in to reduce lateral forces, I don't believe you want to go as far as balancing it out completely: in the black ice case you would actually be forced to always drive at that speed, and a more careful slower driving (or a queue) would be dangerous as you would fall in the corner. That's not the kind of non-linear behaviour you want to have your average driver to cope with.

Nope. Check it out here: www.learning.glass Cheers, Dr. A

Same

Tbh our education system is fucked up