RoyalXBlood Posted October 23, 2008 Share Posted October 23, 2008 The problem: A concrete highway curve of 80 meters is banked at a 19 degree angle. What is the maximum speed with which a 1800kg rubber-tired car take this curve without sliding? (static coefficient of friction of rubber on concrete = 1) So would I set up something like this? The answer is 40.1m/s, but I get 31.6m/s. So I need help to figure out what I'm not getting. I have the force of gravity in the y direction equal to 16,679.0N and in the x direction it is equal to 5,743.0N. The max friction would equal the normal force since the coefficient is 1, so the max friction would equal the force of gravity in the y direction. Add the max friction and Fgx and I have 22,422.0N. That means I can have the force in the opposite direction before sliding, right? If so F=ma or F=m(v^2)/r so 22,422.0=(1800/80)v^2. That's how I got v=31.6m/s Link to comment Share on other sites More sharing options...
swansont Posted October 23, 2008 Share Posted October 23, 2008 22422 N is the force in the x direction of your coordinate system. That's fine. What direction is the centripetal force in your coordinate system? Link to comment Share on other sites More sharing options...
RoyalXBlood Posted October 24, 2008 Author Share Posted October 24, 2008 Would it be in the same direction as the 22422N? If so I wouldn't know what to do from there.. Link to comment Share on other sites More sharing options...
swansont Posted October 24, 2008 Share Posted October 24, 2008 Would it be in the same direction as the 22422N? If so I wouldn't know what to do from there.. The centripetal force is the force directed toward the center of the circle However, I think you have to revisit the approach to the problem. I've looked at it a little more carefully; the net force is not zero, which changes how you have to look at the free-body diagram, and the normal force is not merely a component of the weight. If you look at the forces present in the coordinate system where gravity is in the -y direction, you'll see that the y component of the normal force is greater than the weight. (the car is, in effect, "pressed down" to the track by going fast, since the track has to push it in a circle. Consider the case where the track is inclined at 90º to see an extreme example of this) Link to comment Share on other sites More sharing options...
RoyalXBlood Posted October 28, 2008 Author Share Posted October 28, 2008 I'm assuming my difference in the final answer comes from miscalculating the correct normal force... how would I calculate it from the given information? Link to comment Share on other sites More sharing options...
swansont Posted October 29, 2008 Share Posted October 29, 2008 I think the easiest way to do that is to use a coordinate system where gravity is in the -y direction. The you should see that the forces in the y direction are gravity (down), a component of friction (down) and a component of the normal force (up). By inspection, you should see that the normal force has to be larger than the weight. Since the frictional force can be expressed in terms of the normal force, it can be solved. Then you solve for the x-component of the force, which is the centripetal force. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now