# Homework Help: Banked curve with friction

1. Jan 26, 2006

### James_fl

Hello there. I need some help in solving the following problem:

Problem: A race-car driver is driving her car at a record-breaking speed of 225 km/h. The first turn on the course is banked at 15 degree, and the car's mass is 1450 kg.
a) Calculate the radius of curvature for this turn.
b) Calculate the centripetal acceleration of the car.
c) If the car maintains a circular track around the curve (does not move up or down the bank), what is the magnitude of the force of static friction?
d) What is the coefficient of static friction necessary to ensure the safety of this turn?

I'm not sure if my answers are correct, since I assume I don't need to take into account the friction forces to calculate a) and b).

For a), I used the formula: r = v^2/(g.tan A). The result is 1487.584 km.

For b), a = (v^2)/r = 2.626 m/s^2.

Now.. I'm stuck in c). From common sense, the force of static friction will be 0 since I am assuming there is no friction to calculate A and B. That unless my assumption is wrong.

Any help to solve this problem is greatly appreciated!

2. Jan 26, 2006

### James_fl

And to calculate the static friction, I did try to breakdown the centripetal force and the gravity force. But the result is also 0.

So it will be like this:

Fs = static friction force
Fg = gravity force
Fc = centripetal force

Fs + Fg sin A = Fc . cos A
Fs = m*a*cos A - m*g*sinA
Fs = m (a*cos A - g*sin A)
Fs = 1450 (2.626 * cos 15 - 9.8 * sin 15)
Fs = 1450 * 0
Fs = 0

Am I doing something wrong here?

3. Jan 27, 2006

### andrevdh

I sympathize with you - whoever thought this problem out should be flogged in public.
I attempted to solve c) by taking the components of the forces in the horizontal (x) and vertical (y) directions (with friction acting down the bank). In the y direction we have no resultant force and in the x direction the resultant of the components give the centripetal force. Then one eliminates the normal force N by dividing the two equations which gives you one equation with the friction as the only unknown.

Last edited: Jan 27, 2006
4. Jan 27, 2006

### maverick6664

Question c and d are quite separate ones, and you can answer quickly 0, because in question a you (are supposed to) have got the right radius with which the car won't slip outside or inside regardless of friction. So questions c and d are given to tell you when friction comes in or does not....I guess.

Doesn't it make sesnse?

If you are given a certain radius, questions c and d will have the "right" answers.

Last edited: Jan 27, 2006
5. Feb 1, 2006

### James_fl

Sorry for the late reply and thanks for the ideas, guys.

But I have some other questions that just came up while I rethink about the concept of centripetal force. A sample of free body diagram can be found in this website: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html#c1 . The centripetal force creates an angle against the slope. Why is it not directed down the slope?

According to Newton's second law, to keep the car in a position, the resultant force acting on it needs to be zero. However, from the FBD, the only forces acting on the car are directed toward the centre of the circle. But as the car is curving, it maintains a constant height or position relative to the centre of circle. From my reasoning, there should be another force acting against the centripetal force. However, the component of friction as shown in that website is directed to the centre. And I cannot think any other force that may act against the centripetal force. If there's no other force, the car will just simply sink to the centre of circle. So what is that other force?

Last edited: Feb 1, 2006
6. Feb 1, 2006

### andrevdh

The car are experiencing only two real forces - its weight $\vec W$ downwards and the reaction force from the surface $\vec R$ as a result of the car being pulled down by its weight onto the surface. If the surface is frictionless the reaction force on the car will be perpendicular outwards from the surface. If there is friction present, which will be parallel to the surface, the reaction force can be seen as consisting of two components, the perpendicular component $\vec N$, and the friction component $\vec f$.

7. Feb 1, 2006

### James_fl

Yes, but it still doesn't explain why the car doesn't fall to the centre of the circle. I think I am lacking some concept of centripetal force. Centripetal force is a force that is directed to the centre of the circle. So in that sense, if there is no force acting against it, the car should fall inward to the centre of the circle. However, we know that the car maintains a constant height and distance from the circle. Could anyone help me understand how it actually happen?

8. Feb 2, 2006

### andrevdh

According to NI the car would move in a straight line unless a force compells it to change its motion - in this case it is moving along a curve, so that leads us to the conclusion that some force is pushing it towards the inside of the curve. We gave this force a special name - the centripetal force - when the object is moving in a circular path. This force is realized by the horizontal (inwards pointing) component of $\vec R$ in the case of the car.

9. Nov 9, 2006

### MattsVai

Someone should be flogged for this question... doesn't really matter who. Lol

10. Nov 9, 2006

### OlderDan

Did you state all the information given in this problem? There is no way of knowing the radius of the curve from the information given. If you assume there is no frictional force to do part a), the later parts of the problem make no sense.

11. Nov 10, 2006

### MattsVai

we can use r=(v^2)/(gtan15)
this will give us the radius.

I am too stumped as to how to calculate for c)

12. Nov 10, 2006

### OlderDan

That equation only applies if there is no frictional force acting. The problem does not say there is no friction in the first place. It then asks you to find the friction, implying that it is present. The problem is either flawed or it has not been correctly stated by the OP.

13. Nov 10, 2006

### MattsVai

Well, from the little I understand it appears that the car's acceleration IS the static friction. Therefore it is impossible to assume no friction because of Newtons 1st Law: the car would continue in a straight line... correct me if I am wrong.

Working on an answer for c)... Someone hand me a STICK!! arrrg!

14. Nov 10, 2006

### OlderDan

A car could round a banked curve even if there were no friction if it were going at exactly the right speed. The equation you wrote earlier is the equation that gives the required reltionship between speed, radius, and bank angle for the no-friction case. If the car were going faster, it would slide up the bank, but not go straight. If it were going slower it would slide down the bank.

If you use the velocity you calculated for a), the answer to c is zero.

Last edited: Nov 10, 2006
15. Nov 10, 2006

### MattsVai

You make a valid point.... but then if Fs = 0 then why ask such a stupid question? Even worse, why ask 2!? lol

16. Nov 10, 2006

### OlderDan

Exactly the point of my first post. The problem as stated is nonsense.

17. Nov 12, 2006

### MattsVai

The equations for Newton's 2nd Law in x and y together with the relationship between normal force and static friction give three constraints on the unknowns of the problem. It is not possible to solve a problem with four unknowns with only three constraints. One of the unknowns must be specified a priori, or we need an additional constraint.
In my professional opinion, the problem as stated is not solvable unless you _assume_ the answer to any one of the four questions.
<aas58@cornell.edu>
Physics Department
Cornell University

Last edited: Nov 12, 2006
18. Nov 10, 2008

### randomwinner

I believe I know what the answer to your problem is. That is, the problem of not knowing how to start this problem. In truth, it is just a ridiculous hypothesis I have invented, unsure of whether or not this will answer the perplexities that have haunted you.

You are quite correct when you say that the information is insufficient. Nothing could be closer to the truth. the reason for this is that the question never states whether or not the speed stated is the safe maximum speed for this curve./B]

I believe that we must assume this is the safe maximum speed, and therefore can use the formula: v2 = g r tan of theta.

When using this formula in which v is the safe maximum velocity, the information is sufficient. Therefore, the problem can be solved by applying this formula.

If I have made a mistake or am assuming wrongly, do not hesitate to inform me.

19. Jul 13, 2009

perhaps this may work?

Us = Fc/Fn

Fn = mg = 1450kg*9.8m/s^2 = 14210N

Us = 3808N/14210N = 0.267?

Last edited: Jul 13, 2009
20. Jan 9, 2011