Banked Curve- find the Velocity when static MU=0

AI Thread Summary
The discussion revolves around calculating the speed of a rubber-tired car on a banked curve with a radius of 71 m and a bank angle of 12° when the static coefficient of friction is zero. The normal force acting on the car was correctly calculated as 168,971 N. However, the challenge arises in determining the maximum speed without sliding, as a static friction coefficient of zero implies that friction cannot prevent sliding. The interpretation suggests that without friction, the car would slide sideways down the bank, and the question seeks the speed at which the car remains stable on the banked curve. The conclusion emphasizes the need to understand the balance of forces in the absence of friction.
Phoenixtears
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SOLVED

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Homework Statement


A a 14600 kg *rubber-tired car moves on a concrete highway curve of radius 71 m is banked at a 12° angle.

(a) What is the normal force acting on the car?
168971N

(b) What is the speed with which the car can take this curve without sliding? (Assume s = 0.)
m/s


Homework Equations



2nd law statements

a= V^2/ r

The Attempt at a Solution



So I began part a by drawing force diagrams and then 2nd law statements. I was left with:

n= mg/ cos@
n= (14600)/ cos12
n=168971

That was correct. I then attempted to move along to part b. I began this by saying that the total force is static friction, therefore:

Fs(the maximum)= Mass(V(maximum)^2)/r
V^2= (mu)*gr


However, with mu=0 then this all totals to zero. I'm not sure how to get around this. Plus, logically if the coefficient of static friction is zero, then no matter what wouldn't the car slide??

Thanks in advance!

~Phoenix
 
Last edited:
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My only interpretation of this question is this:

- The car will slide DOWN side ways because there is no friction on the road. If he went super-fast (over-exaggeration of course), the car would slide UP because of the effective weight from it's the centrifugal force (not centripetal). So you are asked for the speed at which the car will not slide up or down the bank.

That is my interpretation of the question.

Best Regards,
Sam :smile:
 
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