Circular Curve of a Highway problem

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Homework Help Overview

The problem involves calculating the angle of banking for a circular highway curve designed for vehicles traveling at a specific speed. The radius of the curve is given, and the scenario assumes no friction acting on the vehicles.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the forces acting on the vehicles, questioning the role of friction and exploring the use of trigonometric functions to find the angle of banking.

Discussion Status

Some participants have provided hints regarding the use of tangent functions to approach the problem, while others have confirmed the absence of friction. There is an ongoing exploration of the correct relationships between the forces involved.

Contextual Notes

Participants are navigating assumptions about the forces acting on the vehicles, particularly the implications of having no friction in the scenario. There is also mention of confusion regarding the application of gravitational force in the calculations.

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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?
 
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Hi mossfan563! :smile:
mossfan563 said:
… 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:
 
There is no friction.
 
mossfan563 said:
There is no friction.

i know! :smile:

you mentioned it!

try the tan(theta) thing :smile:
 
tiny-tim said:
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!
 
mossfan563 said:
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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