Circular Curve of a Highway problem

  • Thread starter mossfan563
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  • #1
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Homework Statement


A circular curve of highway is designed for traffic moving at 95 km/h. Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is 110 m, what is the correct angle of banking of the road?


Homework Equations


F_y = N sin(theta) = (mv^2)/r
F_x = N cos(theta) = mg


The Attempt at a Solution



I assumed that if N = mg then I could cancel out m from either equation since I don't know it initially.
Then I would be left with:
sin(theta) = ((mv^2)/r)/N
cos(theta) = (mg)/N

I used my sin equation. With v = 26.38 m/s and r = 110 m and g = 9.8 m/s^2
I ended up with: sin(theta) = .64589 which is the coefficient of friction.
Doing the inverse sin of that equation, I got an angle of 40.23 degrees.
It was incorrect. What am I doing wrong?
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Hi mossfan563! :smile:
… sin(theta) = ((mv^2)/r)/N
cos(theta) = (mg)/N

I used my sin equation. With v = 26.38 m/s and r = 110 m and g = 9.8 m/s^2
I ended up with: sin(theta) = .64589 which is the coefficient of friction.
Doing the inverse sin of that equation, I got an angle of 40.23 degrees.
It was incorrect. What am I doing wrong?
sin(theta) ? friction?? :confused:

and what happened to g?

Hint: tan(theta) = … ? :smile:
 
  • #3
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There is no friction.
 
  • #4
tiny-tim
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There is no friction.
i know!! :smile:

you mentioned it!

try the tan(theta) thing :smile:
 
  • #5
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i know!! :smile:

you mentioned it!

try the tan(theta) thing :smile:
Sorry I thought LowlyPion replied to my question. I got an email with his response asking if there was friction or not.

Thanks for the hint! It worked!
 
  • #6
LowlyPion
Homework Helper
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Sorry I thought LowlyPion replied to my question. I got an email with his response asking if there was friction or not.

Thanks for the hint! It worked!
Sorry. I withdrew my post when I saw someone else was helping you already. Too many cooks and all that. You found the Cosθ term ... so carry on.
 

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