# Determine a formula for the angle at which a road should be banked

1. Nov 16, 2004

### HawKMX2004

Q: For a car traveling with speed 's' around a curve of radius 'r', determine a formula for the angle at which a road should be banked so that no friction is required to keep the car on the road. Then, find the angle for a curve of radius 50 m at a speed of 50 km/h.

I drew a Free Body diagram of an Inclined Plane, and the Circle the car is traveling. The following is my equations

NetForce(Y-direction)=Fn(ForceNatural) - Fgy(ForceGravity*Ydirection*) = 0
Nfy = Fn = Fgy
Nfy = Fn = Fg(Cos(Theta))

Nfc(Netforcecentrifical) = Ff(ForceFriction) = m(v^2)/r
Nfc = FnUs = m(v^2)/r
Nfc = Fg(Cos(Theta))Us = m(v^2)/r
Nfc = ma(cos(Theta))Us = m(v^2)/r mass cancels out
Nfc = a(cos(Theta))Us = v^2/r
Nfc = cos(Theta)Us = v^2/r(a) Us = V^2/r(a)
Nfc = cos(Theta) = (v^2)Us/r(a)
Nfc = Theta = cos^-1[(v^2)Us/r(a)]
Nfc = Thata = cos^-1[v^2(v^2)/ra(ra)]
Nfc = Thata = cos^-1[v^4/r^2(a^2)] <------ Final Equation

Nfc = Thata = cos^-1[50^4/50^2(9.8^2)]
Nfc = Theta = 89.9 Degrees <---- Final Answer

I think i did something terribly wrong, that answer doesnt seem right. Help would be very greatly appreciated...Thankz

2. Nov 16, 2004

### Diane_

I'm afraid I gave up following your solution early on. Not your fault - it's just difficult to read without the standard notation. *sigh* One of these days, I have to take the time to learn to do that.

Anyway - think about the problem. The car is travelling in a circle, which means that there must be a centripetal force directed towards the center of the circle. You have two questions to answer: 1) Where is the center of that circle and 2) What provides the centripetal force?

The answer to 1 is pretty easy: envision the path of the car. It'll make a circle, and the center of the circle is the center for which you're looking. Note that the center will not be on the ground. One mistake some students make is putting the center on the ground in the middle of the curve.

For 2, there's really only one answer as well: the centripetal force is provided by the component of the normal force pointing towards that center. That component will be parallel to the ground. You have to find the banking angle that makes that component equal to the centripetal force necessary to hold the car in the circle.

I hope that helps - if not, post again and tell me where it fails, and we'll try it again.

3. Nov 16, 2004

### HawKMX2004

Re: Help plz

I got that, that was 89.9 degrees, which i find wrong...do you have any other suggestions or equations i might be able to use?

4. Nov 16, 2004

### Diane_

OK - I did the problem quickly and ended up with

theta = Arctan(v^2/rg)

Keeping proper track of the units (remember - you have the speed in kph and you need it in m/s), I end up with an angle of 21.5 degrees.

Part of the key to this is realizing that the component of the normal force perpendicular to the ground has to cancel out the weight.

Does this help any?

5. Nov 17, 2004

### HawKMX2004

Yes it does! I understand it now, somewhere I lost track of what i was doing but yes now it makes sense. Thank you very much!