Banked Roadway Design for Ice Conditions: Solving for Optimal Angle

[scharry03]

Homework Statement

A civil engineer wishes to redesign the curved roadway in Interactive Example 5.7 in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the desig- nated speed can negotiate the curve even when the road is covered with ice. Such a curve is usually banked, meaning that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the curve is to be 13.4 m/s (30.0 mi/h) and the radius of the curve is 35.0 m. At what angle should the curve be banked?

Homework Equations

nsin(theta) = mv² /r
ncos(theta) = mg
tan(theta) = v²/rg

The Attempt at a Solution

I had thought that looking at horizontal movement on the slope you would use mgsin(theta) = mv²/r, so that the force is parallel with the slope. Why are we not doing this? Thanks.

 

[paisiello2]

Wouldn't it depend on how you measured r?

 

[billy_joule]

You can assume r is measured to the COM of the car.

The Attempt at a Solution

I had thought that looking at horizontal movement on the slope you would use mgsin(theta) = mv²/r, so that the force is parallel with the slope. Why are we not doing this? Thanks.

For problems like this it is often easier to define an axis parallel to the slope and work that way. Then you can ignore the normal force as it makes no contribution to what you are interested in; what theta is when all forces parallel to the slope sum to zero.

With that said, the force due to the centripetal acceleration is not parallel to the slope, it's horizontal. so your expression mgsin(theta) = mv²/r will not give the correct answer. Those two force vectors are not collinear.

 

[scharry03]

Okay, I didn't realize that centripetal acceleration wasn't parallel with the slope, but now that I think about it, that makes complete sense. Thanks!

 

[dean barry]

Have a look at the sheet i have attached, note that its an intuitive solution.

 

[haruspex]

Have a look at the sheet i have attached, note that its an intuitive solution.

For neutral banking, the banking will be perpendicular to the force resultant (F3)​

You should clarify that that is using the centrifugal force view, not an inertial frame.