# What is wrong with my derivation of max speed on inclined curves?

1. Jun 4, 2005

### michaelw

For a flat road, where a car is turning, the max speed can by found by:
Fc = Ff
mv^2/r = uFg = umg
v = sqrt(rug)

For a banked road with angle T, the max speed (i think i did this wrong)
mv^2/r = Fnsin(T) + Ffcos(T)
mv^2/r = mg*sin(T) + umg*cos(T)
v = sqrt(rg(sin(T) + ucos(T))

but sin(T) + ucos(T) can be < 1 for low u's! this doesnt make sense.. in other words its saying that in some cases, driving on a bank results in a lower max v than driving on a flat road..

what did i do wrong?

2. Jun 4, 2005

### Staff: Mentor

Your error is in assuming that the normal force equals mg. This is incorrect for a banked road. You'll need a second equation (for the vertical forces) to solve for the normal force and thus the max speed.

3. Jun 4, 2005

### Andrew Mason

As Doc Al has pointed out, the normal force is not mg. It is the component of mg that is perpendicular to the surface + a component a centripetal force. Your first statement of the centripetal force is correct (where Ff = uFn). The trick is to find Fn.

In order to determine the normal force, you have to look at the vertical components of the forces (which sum to ___?). So there is an additional equation that you need in order to work out the normal force.

AM

4. Jun 4, 2005

### michaelw

thanks
the vertical components should add to 0 :)

5. Jun 4, 2005

### Andrew Mason

The friction force has a downward vertical component and this, together with gravity, equals the upward vertical component of the normal force:

$$F_{gy} + F_{fy} = F_{Ny}$$

What is the vertical component of the gravitational force (easy)?

What is the downward vertical component of the friction force? (The friction force is parallel to the surface so it has a horizontal and downward component. The friction force along the surface is $\mu_sF_n$)

What is the upward vertical component of the normal force?

AM

6. Jun 5, 2005

### michaelw

The vertical component equation is
Fy = Fn*cos(T) - Fg - Ff*sin(T) = 0

The horizontal component is
Fx = Fc = Fn*sin(T) + Ff*cos(T) = mv^2/R

but how can i solve these to find an equation for velocity?

7. Jun 5, 2005

### OlderDan

If you are still looking for maximum speed under these conditions (banked curve) the frictional force and the normal force are stll related by the coefficient of friction. You had the wrong normal force in your first post. Now that you have that corrected, you can use the friction equation.

8. Jun 5, 2005

### michaelw

how can i find what Fn is though from my equations ? Do i need to know trig rules?

The vertical component equation is
Fy = Fn*cos(T) - Fg - Ff*sin(T) = 0

The horizontal component is
Fx = Fc = Fn*sin(T) + Ff*cos(T) = mv^2/R

9. Jun 5, 2005

### OlderDan

You find the normal force by solving your first equation using the fact that Ff is proportional to Fn. How are they related?

I am assuming the bank angle and the coefficient of friction are given.

10. Jun 5, 2005

### michaelw

Fy = Fn*cos(T) - Fg - Ff*sin(T) = 0
Fn(cos(T) - u*sin(T) = Fg
Fn = Fg/(cos(T) - u*sin(T))

Is that correct? It seems weird because Fn would be negative for a high theta..

11. Jun 5, 2005

### OlderDan

That is a good observation. Suppose you start with a flat track. Then Fn would have to be Fg, and your answer reduces to that result. As you increase the bank angle, your result says the normal force will increase. That is because your equations assume that as you increase the bank, you are going to go faster and maintain the maximum possible speed. What is going to happen to the normal force before that denominator goes negative? Suppose you made the bank vertical. Have you ever seen those guys on motorcycles riding inside a cylinder? Which direction do you think friction is acting in that case?

12. Jun 6, 2005

### michaelw

ahhhhhh
that makes sense now :)
so just to be sure, in that case, Fn would be entirely the centripetal force, with no component opposing gravity (ie 90 degrees), and Ff would now oppose gravity, with the force directed up