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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!

they're proportional - its proablya year to late lol

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

This is just basic high school level circular motion.

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

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.

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.

the video is flipped so you can read it...

i think we know scoob@@Siggfuggggg2000

+lolzomgz1337 No. Many times when we deal with a block on a ramp, the block accelerates down the ramp. In those cases we rotate the reference frame so that the acceleration is parallel to the x axis. Since the pull of gravity is not parallel or perpendicular to the ramp, the result of rotating the reference frame is that gravity has components in both the x and y directions. However, in this case there is no acceleration along the ramp, rather the acceleration is towards the center of the circle (we assume the car travels at a constant speed, without drifting up or down the incline). The math is simpler if the x axis is parallel to the acceleration so we do not rotate the frame of reference. This means gravity is parallel to the y axis, so there is no x component of weight.