I have been a highway engineer and have held traffic, transportation, and civil engineer licenses for over 45 years. I worked with the department of transportation for 40 years, please do not forget super elevation of the curves, Some time you encounter with Broken back curves or reversed curves. Thanks
How do u know which way is the friction? from what I can imagine, couldn't the friction also be the other way around ( upward the slope ) like how u can stick something on a slope and it stays at the same spot. Because the friction is resisting gravity making the object not ride down. Or is the bank curve scenario different from the incline example?
mg is vertically downward, so there's no need to resolve it to x and y components here. we can straightaway consider it in the y direction. if resolved however, mgsin(theta) will be along the inclined surface
@@topiary5650 That's not true. Racing tires have coefficients well over 1. There is nothing wrong with friction coefficients over 1. The deal with the model is that it breaks down when Cos(Q)=muSin(Q). Remember the restraint in the model was looking for Vmax, so it is saying that Vmax tends to infinity... in other words you can go as fast as you want and you will never slide out. Look at the 2nd line under Sigma Fy... if Cos(Q)=muSin(Q) then -mg = 0. That's impossible, unless Fn can go to infinity
@@brianmcelhenny7645 I'm not talking about real life I'm talking about what's gonna be on the test. I'm just trying to check off all the boxes for the pre-med requirements over here and they require us to do physics. I appreciate the input though, I do enjoy learning physics and natural sciences and stuff. I'm doing a joint biology and applied math degree cuz of that.
The static force of friction "fₛ" is often defined as "fₛ = μₛ⋅n", which means that it is the product of the static coefficient "μₛ" multiplied by the normal force "n". The static coefficient has different values depending on which surfaces that make contact with each other, and it will usually be given to you.
Hi, How can i produce BACK EMF or voltage spikes in a circuit.. I am able to create voltage drop noise but i need Spikes.. Would be awesome if you help me out and I will mention you on my channel. Thanks
Essentially because there is a reaction force from the road's surface as you drive around the corner. If you were at rest on the incline then the normal force would be as you stated, but because you are moving into a curve, you are pushing into the road, so it pushes back. The extreme case is a vertical wall. Imagine a hockey puck moving along the wall of an ice rink; as it enters the curve the wall pushes in on the puck towards the center of the curve, causing the puck to follow the wall. The normal force causes the centripetal acceleration. The same thing happens with a banked curve, the road surface is in effect a wall that you are driving into.
It is good but review on where you placed friction,it has to be on the contact interface of the car and surface not at the center,only normal reaction and weight are acting at the centre not friction
Actually the normal force would also be acting at the contact interface of the car and surface. But theoretically the effect shouldn't matter if the car is considered a rigid body. Truth is it is not exactly rigid since there is sway between tires and car body on sharp turns.
this is one of the most clear explanations I have found to help my students understand banked curves with friction. thank you
This is the best work I can find on "with friction" situation. Thanks for being so clear!
Thanks! This was the only video I could find that involved friction in the equations, and it is very well explained.
woah he can write backwards
the video is flipped so you can read it...
i think we know scoob@@Siggfuggggg2000
Only video I found that showed how to do this kind of problem correctly! Thank you!!
THANK YOU SO MUCH this video was a blessing
Amazing lesson man. Thanks for the knowledge!
I have been a highway engineer and have held traffic, transportation, and civil engineer licenses for over 45 years. I worked with the department of transportation for 40 years, please do not forget super elevation of the curves, Some time you encounter with Broken back curves or reversed curves. Thanks
This is just basic high school level
circular motion.
U are forever my savior in physics
super helpful and very clear! thanks!
you da homie for this 😎
Very clear, thank you
good stuff helped a lot
great video thanks man
How do u know which way is the friction? from what I can imagine, couldn't the friction also be the other way around ( upward the slope ) like how u can stick something on a slope and it stays at the same spot. Because the friction is resisting gravity making the object not ride down. Or is the bank curve scenario different from the incline example?
Good job bro
My professor never taught us this only to find the friction in a banked curve never to actually apply it so damn is all i gotta say
Thank you sir, thanks to you my hair remains in my head.
For an inclined plane we also have mgsin(theta) which points toward the center of the curve. Why is this not included as yet a 3rd centripetal force?
no it points along the inclined plane
mg is vertically downward, so there's no need to resolve it to x and y components here. we can straightaway consider it in the y direction.
if resolved however, mgsin(theta) will be along the inclined surface
Your board is v cool
Hi... What happens if the angle is 45 degrees and the coefficient of friction is 1? Looks like the denominator under the radical becomes zero?
A bit late on this but the coefficient is always less than 1 or else it doesn't make sense.
@@topiary5650 That's not true. Racing tires have coefficients well over 1. There is nothing wrong with friction coefficients over 1. The deal with the model is that it breaks down when Cos(Q)=muSin(Q). Remember the restraint in the model was looking for Vmax, so it is saying that Vmax tends to infinity... in other words you can go as fast as you want and you will never slide out. Look at the 2nd line under Sigma Fy... if Cos(Q)=muSin(Q) then -mg = 0. That's impossible, unless Fn can go to infinity
@@brianmcelhenny7645 I'm not talking about real life I'm talking about what's gonna be on the test. I'm just trying to check off all the boxes for the pre-med requirements over here and they require us to do physics. I appreciate the input though, I do enjoy learning physics and natural sciences and stuff. I'm doing a joint biology and applied math degree cuz of that.
Why is the Fs equal to the normal force?
they're proportional - its proablya year to late lol
how do i find force of friction?
The static force of friction "fₛ" is often defined as "fₛ = μₛ⋅n", which means that it is the product of the static coefficient "μₛ" multiplied by the normal force "n".
The static coefficient has different values depending on which surfaces that make contact with each other, and it will usually be given to you.
Hi,
How can i produce BACK EMF or voltage spikes in a circuit.. I am able to create voltage drop noise but i need Spikes.. Would be awesome if you help me out and I will mention you on my channel.
Thanks
Is anyone able to help me? Why normal force isn't just equal to to the cosine*m*g
Essentially because there is a reaction force from the road's surface as you drive around the corner. If you were at rest on the incline then the normal force would be as you stated, but because you are moving into a curve, you are pushing into the road, so it pushes back. The extreme case is a vertical wall. Imagine a hockey puck moving along the wall of an ice rink; as it enters the curve the wall pushes in on the puck towards the center of the curve, causing the puck to follow the wall. The normal force causes the centripetal acceleration. The same thing happens with a banked curve, the road surface is in effect a wall that you are driving into.
Thank you very much!
It is good but review on where you placed friction,it has to be on the contact interface of the car and surface not at the center,only normal reaction and weight are acting at the centre not friction
it applies at the center of mass, this is why he puts it in the center
Actually the normal force would also be acting at the contact interface of the car and surface. But theoretically the effect shouldn't matter if the car is considered a rigid body. Truth is it is not exactly rigid since there is sway between tires and car body on sharp turns.
well yes but its simplified at this level