# Banked Curve (its killing me)

1. Oct 11, 2005

### Seiya

Hey lads, i solved part a of this problem in 5 minutes, part B is killing me, ive been doing it for 1 hour and nothing.. ive tried everything, any help appreciated:

[Tipler5 5.P.090.] A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when the coefficient of static friction between the road and rubber is 0.05, a car at rest must not slide into the ditch and a car traveling less than 50 km/h must not skid to the outside of the curve. What is the minimum radius of curvature of the curve and at what angle should the road be banked?
____??? m
2.862° (i found this and its correct)

i really dont know if im thinking right anymore after 1 hour of this + other stress that came on so well here is the last thing i tried....

i set my set of axis parallel to the banked road, so the positive Y is the normal force.... and the friction force goes parallel the banked road....

i set the force friction going up the hill to balance the weight component going down the banked road and i had a force component (mv^2/r)/(cos(2.862)

Well i dont even know what im doing anymore.. i know at first i must of been close to solving it but now im too tired to reason ... help appreciated, thanks!

2. Oct 11, 2005

### mukundpa

If the car is at the werge of sling away, the friction must be down the incline.

3. Oct 11, 2005

### Cyrus

No, friction acts up the incline mukundpa. Its what holds the car up from falling. That would amount to saying that friction forces the car down the incline.

Edit: I did not read the whole question, were both right. You have two conditions. If the car slides down into the ditch, friction acts up. If the car goes to fast and slides up the road, friction acts down.

Last edited: Oct 11, 2005
4. Oct 12, 2005

### mukundpa

hence friction must be down the incline.

5. Oct 12, 2005

### Cyrus

6. Oct 12, 2005

### stmoe

I would just use like...
Normal Force in x (component acting against the car skidding) = mass * v^2 / r
F_n sin(theta) = (mv^2) / r
F_n sin(theta) is just F_nx, so F_nx = mv^2 / r

In my first physics class we were told to assume an object moving on ice as moving on a frictionless (perfect) surface ... if the same for you then ignore this part
Forces holding car = Kin. Friction force in x + Normal force in x
F_hc = F_kx + F_nx
F_kx = F_n*u_k * cos(2.862 degrees)
F_hc is the net force thats 'holding' the car on the road ... and ignore the F_nx is the only force thats 'holding' the car ,

The force to hold the car on the road can be seen as...
F_n sin(2.862) + F_n(u_k) cos(2.862) = m (v^2) / r
v, in m/s is 13.888888888... m/s
so
r = m(13.8888888888)^2 / (F_n (sin(2.862)+u_k cos(2.862))

-- if you were ignoring, stop here --
Since F_nx is the only force thats 'holding' the car on the road, then it is your net force that you hve to 'stabilize' the car ... so, we can pull the car in on a tighter radius until the forces wanting to push it off the road overcome the forces wanting to let it 'slide' into the ditch ... no change in y thus the reason only x ..

Because ....
F_n sin(2.862) = mv^2 / r ,
...we can write...
r = mv^2 / (F_n * sin(2.862) )
v is in km / hr, so in m /s its 13.8888888 ... or just (13 + 8/9) m /s (fractions are exact for calculations...)

Hope this helped... If you need a diagram or something then I'll 'toss ' one together in Gimp or something. Once again: I might be totally wrong.. if so then just ignore me.

7. Oct 13, 2005

### Seiya

i solved part 1 lads , i tought i stated it :(

Stmoe: you use the x component of the normal force... i used the x component of the weight, same thing ..

i think i know where my mistake was..

i set mv^2/r towards the middle... according to what i hear you guys are saying that mv^2/r acts up the incline and the friction and the weight x component act down? If so here is where my mistake lay... thanks for helping me realize that!

Last edited: Oct 13, 2005
8. Oct 13, 2005

### mukundpa

Sorry, I was thinking of the second part in which the car is moving with maximum posibble welocity.

9. Jul 13, 2006

### strongwiththeforce

you need to treat it as two seperate questions. one where the point of reference is paralell to the (presumed) ground, that is for the moving one. and then for the question where the car is still the frame of reference would be paralell to the road surface. then you can calculate out the forces acting upon stuff, do alot of cancelling and you should get your answer.

10. Jun 30, 2007

### inam2u

i think iv got the answer

F=(static friction0*R
F=0.05*mg
a=mv^2/r

0.05*mg=mv^2/r
r=mv^2/0.05*mg

r=v^2/g*0.05
r=380.88m.

tan(theta)=v^2/rg
(theta)=2.862
:surprised