Banked Corner w/ Friction Find mu

[Xittenn]

So I had a homework problem and I'm not sure where I keep making the mistake, the answer is 0.36 and I keep getting 0.46!

 

If a curve with a radius of 90m is properly banked for a car traveling 63km/h, what must be the coefficient of static friction for a car not to skid when traveling at 96km/h?

 

[math] \nu_B = 17.5 m/s [/math]

 

 

[math] \nu = 26. \bar 6 m/s [/math]

 

 

[math] \tan \theta = \frac {\nu_B^2}{rg} [/math]

 

 

[math] \theta = 19.13003772 [/math]

 

 

[math] m \frac{\nu^2}{r} = \frac{mg}{\cos \theta} \sin \theta + \mu_s \frac{mg}{\cos \theta} \cos \theta [/math]

 

 

[math] \frac{\nu^2}{r} = g \tan \theta + \mu_s g [/math]

 

 

[math] \mu_s = \frac{\nu^2}{rg} - \tan \theta = \frac{(26. \bar 6)^2}{90 \cdot 9.81} - \tan 19.13003772 = 0.46 [/math]

 

 

What am I doing wrong???

[DrRocket]

So I had a homework problem and I'm not sure where I keep making the mistake, the answer is 0.36 and I keep getting 0.46!

 

If a curve with a radius of 90m is properly banked for a car traveling 63km/h, what must be the coefficient of static friction for a car not to skid when traveling at 96km/h?

 

 

Without working the problem through in detail I cannot find the precise mistake, but it looks to me like you are relying on pre-derived formulas.

 

The basic idea is as follows:

 

1. For a "properly banked" track at 63 km/h the bank angle and radius of curvature are such that gravitational and centripetal forces are precisely matched and the car would stay on the track with a coefficient of friction of 0. This will determine the bank angle, since the radius of curvature is given.

 

2. Now with the bank angle as determined above you re-analyze the situation for a car traveling at 90 km/s and assume a condition of impending slip -- so that the centripetal force, the gravitational force and the frictional force precisely cancel. That will allow you to solve for the coefficient of friction. Note that in this case that the frictional force prevents the car from sliding up the incline. You should also be able to determine, with that same coefficient of friction, the minimal speed allowed to prevent the car from sliding down the incline.

[Xittenn]

Thanks DrRocket! I was the one who derived the formulas, although [math] \tan \theta = \frac {\nu_B^2}{rg} [/math] was in my text and the derivation was sound. Curious that you say the frictional force would prevent the car from slipping up the track. I am solving for [math] \mu [/math] as the component of friction as it slides along the plane parallel to earths surface, in x; at least this is how I was solving it.

 

If the force centripetal is acting on the car outwardly then I am assuming the following:

 

1) the component of the normal force is acting inwardly as [math] F_N \sin \theta [/math]

 

2) the component of the force of static friction acting against the sliding of the car outwardly is [math] \mu_s F_N \cos \theta [/math] where the normal is acting perpendicular to earths surface as the sliding friction is parallel.

 

I did it this way because I assumed that F_c works parallel to earths surface, and so the normal component acting parallel to earths surface in addition to the force of friction acting against the sliding of the vehicle in the direction parallel to earths surface, will cancel.

 

as in:

 

[math] F_c = F_N \sin \theta + \mu_s F_N \cos \theta [/math]

 

I don't see the logic in applying the force of friction in the upward direction or along the slant as the y components of force should cancel to 0 against the force of gravity not centripetal acceleration??? (this last bit is not worded right but there should be no component of static friction acting in the y, I think)

Edited February 15, 2012 by Xittenn