# Physics Giancoli Ch 5 # 21 Determining mu of banked curve

1. Oct 28, 2007

### NewbiNewton

If a curve with a radius of 88m is perfectly banked for a car traveling 75km/h, what must be the coefficient of static friction for a car not to skid when traveling at 95km/h?

Relevent equations: F=ma, Fr=muFN

FNsintheta=m(v2/radius) = mg/costheta x sintheta = m(v2/r) = mgtantheta = m(v2/r) = tantheta = v2/rg

I got as far as determining the angle at 75km/h(20.8m/s), which is about 26.4 degrees. I tried to go further and solve the components of the forces of FN and mg, but I had some difficulties, and even if I had done those correctly, I didn't know how to solve the components of the frictional force. I did see this problem fully worked out on another website, but I could not follow the approach. I was confused by certain things they did. Such as giving a vertical component to friction, for which I thought has no vertical component. I would greatly appreciate it if the problem was worked out for me, and then I can ask questions about the steps that were taken. I have been trying to work this problem out for a couple of days, so I'm looking forward to understanding it finally.

2. Oct 28, 2007

### rocomath

you're approach to find the angle at 75 km/h is correct. did you draw a 2nd force body-diagram for which static friction comes into play?

further: take this problem for example

A curve with a 120m radius on a level road is banked at the correct angle for a speed of 20m/s (1st picture). If an automobile rounds this curve at 30m/s (2nd picture), what is the minimum coefficient of static friction needed between tires and road to prevent skidding?

http://i110.photobucket.com/albums/n104/rox1co/bankedcurve.jpg

Last edited: Oct 28, 2007
3. Oct 28, 2007

### NewbiNewton

I've seen the diagram that takes into account static friction. Since I've seen the problem worked out before. But it was on another website, and I wasn't able to ask questions on there. The second diagram didn't help me though, because it didn't help to clarify why friction had a vertical component to it.

4. Oct 28, 2007

### rocomath

well it shouldn't be vertical, it should be downwards from what is shown in my picture

once you get all the forces, solve for N in your Y and plug that into your X so solve for static friction

5. Oct 28, 2007

### NewbiNewton

Well, here, perhaps this help clarify why I'm lost. Here are the equations that were used by the website I found the problem on.

mgsin0 + F(friction)cos0=maR

FNsin0 + muFNcos0=m(v2/r)

How are these two equations related? They both appear to give you the centripetal force.

What general equation are these two equations going off of? Where did mu come from? Why would you add vertical and horizontal components together like that?

FN=(mv2/r/(sin0+mucos0)

Y component: FNcos0-F(friction)sin0=mg Or FN=mg/(cos0-musin0)

6. Oct 28, 2007

### rocomath

ok take this problem for example and use it to solve your problem.

A curve with a 120m radius on a level road is banked at the correct angle for a speed of 20m/s (1st picture). If an automobile rounds this curve at 30m/s (2nd picture), what is the minimum coefficient of static friction needed between tires and road to prevent skidding?

http://i110.photobucket.com/albums/n104/rox1co/bankedcurve.jpg

7. Oct 28, 2007

### NewbiNewton

I'm afraid I don't understand how you worked your problem either. Your using the same equations that I just said I didn't understand. Also, I didn't get why you were adding 90 and 180 to theta. I'd really appreciate it if my questions about the equations I listed were answered because I feel I can solve the problem once I understand the equation.

8. Oct 28, 2007

### rocomath

sorry i went to study at BnN, but i can answer your question now that i know what you're confused about.

ok so take a look at the FBD

looking at the forces in the X-1, we have Normal force and Acceleration (which is replaced by $$a_{rad}=\frac{v^{2}}{R}$$)

i could have assigned trig identities to both of my forces

$$N\cos{(90+\theta)}=ma\sin{180}$$

but i don't assign a trig identity to "ma" bc sin180 is simply -1, so it saves time by already computing it. whenever forces lie on an axis, you don't have to bother with assigning a trig identity to it. (but i think you already know this part, just saying it incase)

the reason why i was adding theta to both 90 and 180, is b/c the angle that is going to be found will not be the right angle. i think it's safe for me to assume that both our problems are similar and the only thing that is different are the numbers.

also, i used the Sum/Differences Trig Identities to simplify my equation so that i wouldn't have to keep writing 90+theta, but you could keep writing it if you'd like since it makes no difference.

the angle that you found was 26.4, but looking at the FBD, that is not correct. b/c Normal force lies within the 2nd quadrant; hence, 26.4 is not in the 2nd quadrant. in order to get the right angle, we must add 90 to that angle (which is theta).

same goes for our 2nd FBD, when we add in static friction, we must had 180 to 26.4.

Last edited: Oct 28, 2007
9. Oct 28, 2007

### NewbiNewton

I kind of understand your method of adding 90 and 180, etc. I've never been taught to use your method with the quadrants, etc. About the angle I determined, I think it is in the first quadrant based on my free body diagram. Your drew yours in the opposite direction that I drew mine. So, I didn't need to add 90 or 180 to theta. The method that is taught at my school is the one I presented in my 4:51 P.M. post. I wish to understand those equations, which are similar to the ones you're using, but with less to consider I feel.

10. Oct 28, 2007

### rocomath

could you link me to it? i don't mind analyzing and explaining it to you if i can.

i'm assuming that you drew your car curving to the left while acceleration was pointing in the positive x-axis.