What is the correct banking for a 580m radius highway curve?

[SherBear]

Homework Statement

A highway curve of radius 580m is designed for traffic moving at a speed of 70.0 km/hr.

Homework Equations

What is the correcting banking of the road?

The Attempt at a Solution

I tried using tanθ=v^2/rg
Also I converted the 70 km/hr into m/s by 70 km/hr(1000 m/km/3600 s/h)=19.44 m/s
So I ended up with 19.44m/s^2/580m(9.8m/s^2)=.066°
It's wrong

Any help is appreciated!

 

[SherBear]

What is the correcting banking of the road?
I mean Correct*---Sorry

 

[LawrenceC]

Draw a free body diagram of the car on the slope and look at the forces on the wheels. Then redo your force balance and work out completely.

 

[SherBear]

I only see normal up, mg down and Ff i guess opposite of the direction, but it doesn't help me =/

 

[LawrenceC]

The vehicle is traveling in a circular path while it is on the banking. Does this suggest another force is acting on the vehicle.

If you look at forces on the wheels you have the normal force you mentioned, the mg force which is the weight of the vehicle. Now suppose the car were stopped and the road were covered with ice. What would the car want to do? What tends to balance this due to the motion of the vehicle?

 

[SherBear]

The car would want to slip and some friction force would cause it not to slip?

 

[LawrenceC]

The car would want to slip and some friction force would cause it not to slip?

Ok, I was checking your reasoning.

I tried using tanθ=v^2/rg This is correct.


19.44m/s^2/580m(9.8m/s^2)=.066°

I would write this as:
tan(theta) = ((19.44)^2)/(580 * 9.8) = 0.066

What is theta?

 

[SherBear]

Ok, I was checking your reasoning.

I tried using tanθ=v^2/rg This is correct.


19.44m/s^2/580m(9.8m/s^2)=.066°

I would write this as:
tan(theta) = ((19.44)^2)/(580 * 9.8) = 0.066

What is theta?

No idea? but the answer from masterphysics is 3.80°, how do you get that?

 

[LawrenceC]

No idea? but the answer from masterphysics is 3.80°, how do you get that?

You did the hard part of the problem!

tan(theta) = ((19.44)^2)/(580 * 9.8) = 0.066

Therefore we seek the angle such that the tangent of the angle is 0.066.

theta = arctan(0.066)

That angle is 3.8 degrees. What's the problem?

 

[SherBear]

You did the hard part of the problem!

tan(theta) = ((19.44)^2)/(580 * 9.8) = 0.066

Therefore we seek the angle such that the tangent of the angle is 0.066.

theta = arctan(0.066)

That angle is 3.8 degrees. What's the problem?

[/I]

I'm not that smart and I don't know how to do it on my calc? lol

 

[LawrenceC]

Do you see the abbreviation 'tan'? Do you have an inverse button? Mine has a tan button and an inverse button. Another caluclator of mine has a second function button that makes the tan button tan^-1 which is the inverse. These calculators are all different. Use your instruction book.

 

[SherBear]

Geeze, now my calculator wants to work after I got the question wrong, tan-1(.066)=3.8 deg.

 

[SherBear]

Thank you for all your help Lawrence! Now I understand it :)