# Homework Help: Banked Roadway Physics

1. Jun 13, 2015

### scharry03

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
nsin(theta) = mv2 /r
ncos(theta) = mg
tan(theta) = v2/rg

3. 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.

2. Jun 13, 2015

### paisiello2

Wouldn't it depend on how you measured r?

3. Jun 13, 2015

### billy_joule

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

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.

4. Jun 14, 2015

### 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!

5. Jun 15, 2015

### dean barry

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

#### Attached Files:

• ###### nuetral banking.docx
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6. Jun 16, 2015

### haruspex

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