Deriving the Formula for Banked Roads: No Friction

[Brute]

Homework Statement

Derive the formula to calculate the angle at which the road should be banked to prevent the car from skidding(Ignore force of friction).

Homework Equations

tan x = v^2/gr ----- answer
mv^2/r = fc

The Attempt at a Solution

I get how they derived the formula due to this website http://batesvilleinschools.com/physics/phynet/mechanics/circular%20motion/banked_no_friction.htm

My question is how can I do it, by putting mg in terms of horizontal and vertical components, instead of Fn.(I know the website explains why it makes more sense to do it that way, but I still want to know).

My attempt:

Forces in Y = 0:
Fn - mg cos x = 0
Fn = mg cos x

Forces in X:
mg sin x = mv^2/r
g sin x = v^2/r
sin x = v^2 / gr --stuck not the same answer

If someone can please explain what I am doing wrong much help appreciated.

 

[TSny]

Hi, Brute. Welcome to PF!

Forces in Y = 0:
Fn - mg cos x = 0

If the Y axis is perpendicular to the plane (road), then the sum of the forces in the Y direction will not be zero. The car is going in a circle at constant speed. Think about the direction of the acceleration of the car and then think about whether or not that acceleration has a non-zero Y-component.

Forces in X:
mg sin x = mv^2/r

Is v²/r the X-component of acceleration?

 

[Brute]

Hi, Brute. Welcome to PF!

If the Y axis is perpendicular to the plane (road), then the sum of the forces in the Y direction will not be zero. The car is going in a circle at constant speed. Think about the direction of the acceleration of the car and then think about whether or not that acceleration has a non-zero Y-component.Is v²/r the X-component of acceleration?

Im not sure if I understand you correctly. But what i got is centripetal acceleration goes towards the circle, making it not the X-component of acc. But(assuming I understood right) it goes directly opposite the Fn force.

So after I drew a new diagram I got:

Fn = mv^2/r + mg cos x

Fx = 0

mg sin x = 0

But I am totally stuck after that.(Thank you for the quick response)

 

[TSny]

Im not sure if I understand you correctly. But what i got is centripetal acceleration goes towards the circle, making it not the X-component of acc. But(assuming I understood right) it goes directly opposite the Fn force.

Yes, the acceleration is toward the center of the circle. How is the direction of "towards the center" related to the direction of your X and Y axes? (Towards the center is not directly opposite Fn.)

 

[Brute]

Im so confused now, been working on it for over 2 hours almost in total. I'm assuming that there is a X and Y component of centripetal force that I don't understand, but I'm unsure of how I would draw it. I also realize that Fnet is equal to mv^2/r but I don't know what to do now. If you could assist me a bit more it would I would be very thankful.

 

[TSny]

In the attached figure, imagine drawing the acceleration vector of the car. Which way would it point? What angle does the acceleration make to the X axis? How would you find the X and Y components of the acceleration?

 

[Brute]

Hopefully I go this right.

For the forces in Y:

Fn = mv^2/r sin X + mg cos X

For forces in X:

Fn Tan x = mv^2/ r cos x + mg sin x

 

[TSny]

Hopefully I go this right.

For the forces in Y:

Fn = mv^2/r sin X + mg cos X

That looks correct.

For forces in X:

Fn Tan x = mv^2/ r cos x + mg sin x

How did you get the term on the left? Does F_n have an X component?

 

[Brute]

Ohh yeah I see it now so
For forces in the x:

mg sin x = mv^2/r cos x

 

[TSny]

Yes, that's right. If you use this equation to find the angle in terms of the speed, does it give the correct answer?

 

[Brute]

OMG your a genius, it works perfectly. Thanks for the all the help man, this made me understand a lot of other problems too.

 

[TSny]

Good. As a general rule of thumb, if you're applying Newton's law to an object and if you know the direction of the acceleration, it's best to choose one of the coordinate axes along the direction of the acceleration. Then you only have one nonzero component of acceleration and that usually makes working with the component equations easier. But as you've shown, you'll still be able to get the correct answer even if you orient your axes some other way.