Banked Turn Equation Discrepancy

[Neptulin]

I know that Wikipedia isn't the best source, but it was what I could find on the web to check my solution.

I have been trying to calculate the maximum velocity for a car going around a banked turn. What I found online, contradicted my final equation.

The following wikipedia page gives a solution to finding the maximum velocity of a car on a banked turn:
http://en.wikipedia.org/wiki/Banked_turn" .
It ends up with this formula:
:<math>v= {\sqrt{rg\left(\sin \theta +\mu_s \cos \theta \right)\over \cos \theta -\mu_s \sin \theta }}</math>
Edit: I don't know how to post an equation in these forums. For the moment you'll just have to go onto the page and scroll down.

When I tried the solution I got almost the same thing, but the sinθ and the cosθ (not the ones multiplied by μ), in the numerator and denominator respectively for Wikipedia, were in opposite positions in my equation. I have checked and rechecked what I've done, but I think the the Wikipedia page was mistaken in the first step of its solution:
"Once again, there is no motion in the vertical direction, allowing us to set all opposing vertical forces equal to one another. These forces include the vertical component of the normal force pointing upwards and both the car's weight and vertical component of friction pointing downwards:
Ncosθ = μsNsinθ + mg"

It also uses Nsinθ in the expression for horizontal motion.

I think that whoever wrote this article mixed up their horizontal and vertical components for the normal force. If θ is measured from the horizontal, then the vertical component, by what I see, has to be represented by sin (it is always opposite). My questions is this, which is (if either) right? And if it isn't me, then why? Thanks.

 

[Doc Al]

"Once again, there is no motion in the vertical direction, allowing us to set all opposing vertical forces equal to one another. These forces include the vertical component of the normal force pointing upwards and both the car's weight and vertical component of friction pointing downwards:
Ncosθ = μsNsinθ + mg"

I don't see a problem. θ is the angle of the road with respect to the horizontal, thus the normal force makes an angle θ with the vertical.

It also uses Nsinθ in the expression for horizontal motion.

Makes sense to me.

 

[Neptulin]

thus the normal force makes an angle θ with the vertical.

You know when you spend ages figuring out how to do something, you are told its wrong and spend ages trying to figure out why - and when you ask for help it becomes obvious you messed up the simplest step, that's how I feel.

I drew a vector diagram with the normal and its components separate from the main diagram, and drew θ from the horizontal for the normal, essentially transferring across without thinking. Well, at least time I'll be mindful of this. Thanks for your help.

And now to slink away in shame.

 

[Doc Al]

No shame required. Happens to the best of us.

 

[Stephen Tashi]

To post LaTex math formulas on the forum, use the tag "tex" in square brackets instead of the tag "<math>".
Or if you are typing a sentence and want the formula to stay in the same line as your words, use the tag "itex".