I would just like to thank you and your channel. I couldn't have survived my engineering course without your videos. Keep on making lectures like this Sir.
I was still confused as to why these kinds of problem precedes my electronics technology classes in community college. In hindsight I am still befuddled at the relationship to an EMV (electromotive force), though a device. I should have noticed the word "force" in the definition.
thank you for this! I've been trying to find out why people were drawing the free body diagram so differently than a regular inclined plane. this helped a lot!
Does the friction force where the rubber meets the road acts radially inward to counter the tangential velocity on the car moving in a circular motion?
Is the friction in the 2nd question, the static friction or kinetic friction? I believe it should be static friction. The only has a tendency to slide up or down and there is no relative motion between the car and the banked plane yet. Unless, this is the maximum static friction and the car is going to slide down or up if the velocity decrease or increase a little bit more. Then we can use static fiction coefficient.
If I had to guess, I’d say it’s the kinetic friction coefficient since we are dealing with a car while it’s driving and we’re studying it’s velocity. However, I could be totally wrong🤷♂️
12:45 one thing I'm not getting, if mv^2/r is brought to the left of the equation it will have a negative sign, however the centripetal force acts in the postive x-direction so why would it be negative there. I understand it it the right answer but I cannot find the reason behind this.
There's no component of weight since weight is perpendicular to the x-axis. Nsinx and mv^2/r are not in the same direction. mv^2/r is an inertial normal acceleration acting in the opposite direction to the force, I'd rather visualize mv^2/r as a result of Nsinx-fcosx since Nsinx and fcosx act opposite to each other in the normal plane i.e x-axis in this particular example.
We don't concern ourselves with components of weight as by convention we have chosen weight to act in the j (y) direction. We could go by more standard convention and choose the direction of normal force to be in the positive unit vector j direction, in which case the problem would be solved differently (a big difference being we now concern ourselves only with the cosine component of centripetal force).
I would just like to thank you and your channel. I couldn't have survived my engineering course without your videos. Keep on making lectures like this Sir.
I was still confused as to why these kinds of problem precedes my electronics technology classes in community college. In hindsight I am still befuddled at the relationship to an EMV (electromotive force), though a device. I should have noticed the word "force" in the definition.
Are you an engineer at present?
thank you for this! I've been trying to find out why people were drawing the free body diagram so differently than a regular inclined plane. this helped a lot!
Definitely underrated video. Thank you!
You are a great explainer dear friend. 👍
I'm here to check on my mans who helped me pass my exams without cheating.
Does the friction force where the rubber meets the road acts radially inward to counter the tangential velocity on the car moving in a circular motion?
yes indeed
Very nice explanation by you...keep it up😊👍
Thanks a lot - the detailed explanation is very much appreciated! You make it so easy to understand!
Is the friction in the 2nd question, the static friction or kinetic friction? I believe it should be static friction. The only has a tendency to slide up or down and there is no relative motion between the car and the banked plane yet. Unless, this is the maximum static friction and the car is going to slide down or up if the velocity decrease or increase a little bit more. Then we can use static fiction coefficient.
If I had to guess, I’d say it’s the kinetic friction coefficient since we are dealing with a car while it’s driving and we’re studying it’s velocity. However, I could be totally wrong🤷♂️
@@DanielRodrigues-oe1bw you are wrong indeed, it's the static friction here between the tyres and the road
Nicely done Patrick. I liked when you assumed mu=0 and showed the problem was like the one you started the video. :)
Wow! thank you so much great explained
12:45 one thing I'm not getting, if mv^2/r is brought to the left of the equation it will have a negative sign, however the centripetal force acts in the postive x-direction so why would it be negative there. I understand it it the right answer but I cannot find the reason behind this.
You really helped me in my assignment.
Thank you
Really well explained, thank you!
Thank you so much! Lots of love from India
wow bro tnx a lot
We missed you sir
Your video r really helpful thank you so much
In the summation of Fx, why is the Nsinx is positive while in the diagram is pointing at negative x-axis? In the banked curve with friction part.
Your video helpfull to me thanku somuch 👏👏
hello sir ,what device you connect to your computer to write on it ? Iam looking to buy one ,I liked the quality of your vedeo
This is useful for IB Physics HL
So good!
Tysm, you saved me from my exam
Good luck! Hope it goes well for you!
Wow this is such an amazing video
Excellent Video. Thank you.
very informative 😍😍
Jacob is pretty awesome!
Omg jeez I just watched a video from 2010 from you and your still going!
Old style is best
the correct term is "superelevated"
Thanks
Thank you so much sir.
Thank you
how do you find the friction??
Friction should be inward to resist centrifugal force
indeed
where are the components of weight? mgsinx and mgcosx, also why isn't it is "Nsinx +mv2/r = fcosx" since nsinx and mv2/r are in the same direction
There's no component of weight since weight is perpendicular to the x-axis. Nsinx and mv^2/r are not in the same direction. mv^2/r is an inertial normal acceleration acting in the opposite direction to the force, I'd rather visualize mv^2/r as a result of Nsinx-fcosx since Nsinx and fcosx act opposite to each other in the normal plane i.e x-axis in this particular example.
We don't concern ourselves with components of weight as by convention we have chosen weight to act in the j (y) direction. We could go by more standard convention and choose the direction of normal force to be in the positive unit vector j direction, in which case the problem would be solved differently (a big difference being we now concern ourselves only with the cosine component of centripetal force).
the best
ANGEL
Sir, if you analys the vector of wieght to its compenents then , the reaction of road N=mg.cos theta. Not mg/cos theta
What type of math is this calc?
circular motion
Algebra and trigonometry used to solve a physics problem.💪👍
we do not use calc here it is algebra and trig
Amazing video. I am a math teacher, can we work together ?
how do I get smart
by not asking such dumb questions
Why u r uploading ur videos soo late?... We need knowledge... And ur videos r very helpful... Show some activeness...
shut up weirdo, he can upload when he wants
when are you going to post helpful physics videos too? show some activeness