# Static friction

1. May 5, 2010

### xstetsonx

A car, travelling at speed v, can safely negotiate an unbanked curve with a 50.0-m radius when the coefficient of friction between the tires and the road is 0.8. How much bank would a curve with the same radius require if the car is to safely go around it at the same speed, v, without relying on friction?

my work:
Msmg=Fr=m((v^2)/r)
(0.8)(9.8m/s^2)=(v^2)/(50m)=19.79m/s

Mgsinx=Fr=m((v^2)/r)
(9.8)sinx=(19.79m/s)/(50m)=53.13 degree

please tell me if i am wrong or right

2. May 5, 2010

### PhanthomJay

Right.
Wrong.

The centripetal acceleration is directed radially (horizontally) inward toward the center of the curve.

3. May 5, 2010

### xstetsonx

so will be
mgsinxcosx=m((v^2)/r)?

4. May 5, 2010

### PhanthomJay

That's getting close, but still incorrect. Please show or indicate how you are arrivig at this equation.

5. May 5, 2010

### xstetsonx

well the only Fr is the gsinx but since you said Fr is directly horizontal to the center so i did cos
but i might be wrong now i am kinda confuse too

6. May 5, 2010

### PhanthomJay

the radial acceleration is not gsin theta. Draw a free body diagram of the car. There are 2 forces acting on the car. One is its weight, acting vertically down. What's the other??

7. May 5, 2010

### xstetsonx

normal force?

8. May 5, 2010

### PhanthomJay

Yes. So of those 2 forces, which one has a component in the x (radial, horizontal) direction? What is that component?

9. May 5, 2010

### xstetsonx

nsinx so it is the same as mgtanx?

10. May 5, 2010

### PhanthomJay

Yes, I'm not sure how you got that, but that is correct.

11. May 5, 2010

### xstetsonx

well i think n=mg/cosx so plug it back into the equation (mg/cosx)(sinx)=mgtanx right?

Last edited: May 5, 2010
12. May 5, 2010

### PhanthomJay

Yes, from Newton 1 in the y direction [/quote]so plug it back into the equation (mg/cosx)(sinx)=mgtanx right?[/QUOTE]yes, solve for theta (or the angle you are calling 'x') = ?

13. May 5, 2010

### xstetsonx

thanks so much otherwise i might get that wrong on the test lol....