Solving Banked Road Problem with Car Speeds of 48 km/h

[Bernie Hunt]

OK, last time I'll bother you guys today. I'm just having a tough time getting things to work out.

A curve of radius 30m is banked so that a car can round the curve at 48km/h even if the road is frictionless. Calculate the banking angle theta for these conditions.

I have;

tan(theta) = v^2 / gr

tan(theta) = 48^2 / (9.8 * 30)

tan(theta) = 7.8367

theta = 82.7 deg

The book has 31 deg.

Any ideas?

Thanks,
Bernie

 

[BishopUser]

48km/h is a dangerous thing to plug into an equation like that. As a habbit always plug the units in with their numbers

 

[Euclid]

You will have to convert 48 km/h to m/s. Or, convert 30m to km and 9.8 m/s^2 to km/hr^2.

 

[Bernie Hunt]

Argh, Thud, thud, thud ...
(The sound of beating my head on the desk again.(

That's the second time I made a units mistake last night. My montra for today will be "Check the units!"

Bernie

 

[jrmed13]

Help!

Okay, so I am doing a similar problem involving a car driving on a banked, circular track (theta=31degrees). I know that to find the centripetal acceleration, I am supposed to say that (mv^2)/r = nsin(theta). Then, I have to solve for n by saying that ncos(theta)=mg. However, I am confused... why can't n=mgcos(theta). My understanding is that two forces are equal in magnitude if the object doesn't move in either direction. The car doesn't move into the road or out of the road... or does it?? please help! I have a test on monday.