# Coefficent of friction

1. Apr 18, 2008

### 72_Cutlass_S

Hello, I'm towards the end of my first semester of Physics 1 (cal based) and I've had this homework problem that has been making want to pull my hair out. I know that the coeffeicent of friction is between 0 and 1 and I have worked out the problem like my professor did but keep getting the wrong answer. Thank you for any help.

Michael

1. The problem statement, all variables and given/known data
A banked circular highway curve is designed for traffic moving at 70 km/h. The radius of the curve is 215 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?

2. Relevant equations
f-mg*sin(x)=ma*cos(x)
N-mg*cos(x)=ma*sin(x)

tan(x)=v^2/Rg

where:
m= mass = cancels out
g= gravity = 9.8 m/s^2
a= acceleration = v^2/R = 11.11^2/215 = .574 m/s^2
x= angle
v= velocity = 40 km/h = 11.11 m/s

3. The attempt at a solution

coefficent of friction = (g*sin(x)-a*cos(x))/(g*cos(x)+a*sin(x))

x= tan^-1(11.11^2/(215*9.8))
x= 3.353

f = (9.8*sin(3.353)-.574*cos(3.353))
(9.8*cos(3.353)-.574*sin(3.353))

f = -1.01*10^-15

2. Apr 18, 2008

### alphysicist

Hi 72_Cutlass_S,

I think there is a problem with your angle. When the problem says that the highway is designed for a speed of 70km/h, that means that at that speed friction is not needed to carry the car around the curve. The formula $\tan\theta = v^2/(rg)$ applies to that special speed. So I think you put in the wrong speed.

There is also a typo in your first equation (f-mg*sin(x)=ma*cos(x)); the right hand side needs a negative sign to be consistent with the left hand side. However, the minus sign appears later on, which is why I think it you just accidently left it out.

The next to the last line also has a minus sign in the denominator which I don't think should be there.