# Find the angle.

• Mar 3rd 2008, 06:04 PM
CenturionMonkey
Find the angle.
I didn't know where else to post this, but I need a lot of help.
I'm stuck on this physics question:

A highway curve in the horizontal plane is banked so that vehicles can proceed safely even if the road is slippery. Determine the proper banking angle for a car traveling at a constant velocity of 97km/h on a curve of radius 450.

I don't even know where to start. Please and thank you.
• Mar 5th 2008, 09:20 AM
sandstorm
Hello,

I car on a slope....

The centripetal force, (mv^2)/r , acts horizontally away from the centre of the bank. The component weight of the car down the slope is mgsin(x).

For the car to not slide up or down the track, this force mgsin(x), must be equally balance by another force which happens to be the component of the centripetal force up the track.

So, mgsin(x) = (mv^2)*cos(x)/r

Then the masses cancel and you can rearrange for the angle, x.

The beauty of this is no matter if you are driving a (frictionless) tractor or riding a bicycle, as long as you are travelling at the critical velocity, you will not slip.
• Mar 5th 2008, 10:02 AM
topsquark
Quote:

Originally Posted by sandstorm
Hello,

I car on a slope....

The centrifugal force, (mv^2)/r , acts horizontally away from the centre of the bank. The component weight of the car down the slope is mgsin(x).

For the car to not slide up or down the track, this force mgsin(x), must be equally balance by another force which happens to be the component of the centripetal force up the track.

So, mgsin(x) = (mv^2)*cos(x)/r

Then the masses cancel and you can rearrange for the angle, x.

The beauty of this is no matter if you are driving a (frictionless) tractor or riding a bicycle, as long as you are travelling at the critical velocity, you will not slip.

You would do better to remove all references to the "centrifugal" force here. So the centripetal force has a magnitude of mv^2/r in the direction of the center of the circle etc. etc.

-Dan