I was wondering, if I were to drive in a flat road , and then on a banked road(steep), in a circular motion, which one would be more difficult to manoeuvre over. I thought of generalising the two, but obviously it won't work because in the flat road, it is the frictional force that acts as the centripetal force(while the normal force cancels the weight), while in the banked road, the horizontal component of the normal force acts as the centripetal force(does anyone what the friction does then?). But anyhow, to measure the 'manoevurabilty', I was thinking of using the velocity, ( more of it, the harder it is to manoeuvre),and so would have used force=mv^2/r, but there was a flaw because like I said the force acting as centripetal are different in each case. So now I am thinking the 'ease with which the car turns in each case' as a measure of difficulty in manoeuvring. This depends on the magnitude of the centripetal force in each case. So i again used force=mv^2/r, this time keeping v constant. So I figured, out that it takes more force in the banked surface than in the flat, to maintain the same speed. Is this a good way to figure it out? or is there a better way? This is not really anything important, and purely a discussive question. I was just thinking bout it, and thought what would by fellow physicsforumers say about it...:) If you guys do have a better way, lets discuss that! and btw Im an A level student and just started centripetal forces. So I know very little as of now. Just so we stay on the same page. I took random variables, which were m of car=5kg, radius=2m, elevation of banked road=30 degrees.