Jump to content

Problem velocity and banked curves. (Physics)


RoyalXBlood

Recommended Posts

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?

phys.jpg

 

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

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

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

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 account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.