Banked Curves using Radius and speed

1. Oct 20, 2009

SlainTemplar

Okay so i have a test and so I'm reviewing for it, and one of the questions i came across was:

1. A curve of radius 180 m is being designed in a new highway to allow cars travelling at 70 km/h to round the curve with zero frictional forces.
a) At what angle must the road be banked?
b) What is the minimum coefficient of friction that will allow a car to successfully go around the curve at 30 km/h?

so i did a, i used Fn = mg/cos theta, and Fn (sin theta) = mv^2/r and substituted for Fn, getting Theta = Tan inverse v^2/gr and i got an angle of 12.1 degrees which is right.

But I'm having a problem getting friction. i tried to assume a mass but that didn't work, and I'm not sure how to start this question. any ideas?

2. Oct 20, 2009

Delphi51

Quite complicated! I find it easiest to take the point of view of the car moving in a circle so you have a centrifugal force pushing the car out. And gravitational force down. You have to take the components of each of those forces into the road and along the road. Use the into the road components to get the normal force and the friction force. Then do the forces along the road surface: the inward/downward component will be just equal to the outward/upward components (including friction).

3. Oct 20, 2009

SlainTemplar

Would i be safe to assume a mass in order to solve for the gravitational force down?

4. Oct 20, 2009

Delphi51

Yes. You can do it for each kg of the car mass. Or just leave the m in each of the force expressions and it will cancel out when you calculate the coefficient of friction.

5. Oct 20, 2009

SlainTemplar

okay i officially hate this problem. im playing with a triangle diagram for Fn and Ff, and im using

Ffx + Fnx = mv^2/r
Ffx = mv^2/r - Fnx

so then i got got a formula for fnx
tan theta = fnx/fny
tan theta = fnx/mg because fny = Fg and Fg = m(9,81)
therefore, Fn = tan theta (mg)

so then i subsituted it into Ffx = mv^2/r - Fnx
giving me Ffx = mv&2/r - tan theta (mg)

im doing all of this because Ff = Ffx/cos theta
so i need ffx to solve for friction, but im not seeing how mass cancels out. am i doing something wrong?

6. Oct 20, 2009

Delphi51

normal force Fn = Fc*sin(12.1) + mg*cos(12.1)
Ff = u*Fn
force up = force down
Fc*cos(12.1) + Ff = mg*sin(12.1)
All the forces have m in them so it cancels out at this point when you fill in the details for Fc and Ff.

7. Oct 21, 2009

SlainTemplar

Well i ended up completing the question, i ended up just going to my teacher for help. i can see why it was so hard for it to be explained xD theres alot u have to do and i was on a totally different track.

Thanks for the help though Delphi51 :)