# Max Speed on an Incline

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1. Mar 9, 2015

### Jacky Lee

1. The problem statement, all variables and given/known data
A concrete highway curve of radius 70 m is banked at a 15° angle. What is the maximum speed with which a 1500 kg rubber tired car can take this curve without sliding? The static friction coefficient is 1.

2. Relevant equations
Centripetal Force = $(m v^2) / r$
Maximum Friction Force = $μN$

3. The attempt at a solution
$(m v^2) / r$ = centripetal force

Friction_max = 1 * normal force = $9.8 * 1500 * cos(15) = 14199.11 N$

Plug back into equation:

$(1500 v^2) / 70 = 14199.11$

Which yields v = 25.7 m/s

However, the answer is 34.5 m/s. What am I doing wrong? I don't understand where I am going wrong in my steps.

2. Mar 9, 2015

### AlephNumbers

Are you sure that the normal force is the only force you have to worry about? No other forces? None at all?

3. Mar 9, 2015

### AlephNumbers

I mean the frictional force

4. Mar 9, 2015

### Jacky Lee

I figured that when the centripetal force exceeds the maximum frictional force, the car would skid. Solving for the maximum frictional force only requires the normal force, so I didn't include any other forces. Where would I have to use the friction force?

5. Mar 9, 2015

### AlephNumbers

Look, I phrased that statement poorly. Reevaluate the forces acting on the car tire.

6. Mar 9, 2015

### AlephNumbers

Everything else is correct, your steps, your concept of friction, all correct.

7. Mar 9, 2015

### Jacky Lee

Forces on the tires: friction, normal, centripetal

Am I missing anything?

8. Mar 9, 2015

### AlephNumbers

Yes. Yes, you are missing something very important. A force that is in every inclined plane problem you have ever seen...

9. Mar 9, 2015

### AlephNumbers

Ignore the centripetal forces. Just think about a tire on an incline. What forces act on it?

10. Mar 9, 2015

### Jacky Lee

OH, I'm missing gravity, aren't I?

11. Mar 9, 2015

### AlephNumbers

I do believe you are.

12. Mar 9, 2015

### Jacky Lee

Hmm, I still don't get the right answer. I tried adding the force of gravity in the direction of the incline to the frictional force (which would just double it), so I got:

$(m v^2) / r = 28398.22$
$(1500 v^2 / 70) = 28398.22$
$v = 36.4$

13. Mar 9, 2015

### AlephNumbers

Don't just give me numbers. Keep it symbolic. What are the two factors that make up the 28398.22N?

14. Mar 9, 2015

### AlephNumbers

The gravitational force directed down the incline would not be cosθmg

15. Mar 9, 2015

### Jacky Lee

I made an error, it should be 18003.75N

Maximum Frictional Force: $9.8 * 1500 * cos(15) = 14199.11$
Force due to gravity: $9.8 * 1500 * sin(15) = 3804.64$
Added together: $14199.11 + 3804.64 = 18003.75 N$

$(m v^2) / r = 18003.75$
$(1500 v^2 / 70) = 18003.75$
$v = 29.0 m/s$

16. Mar 9, 2015

### AlephNumbers

I would wait until you have the solution to start plugging numbers in. You made a sign error.

17. Mar 9, 2015

### Jacky Lee

Should the friction force be negative?

Even if I changed the sign on any of the numbers, my answer would still be incorrect. If I change the sign on either forces, I would get a smaller number and therefore a smaller velocity as an answer.

18. Mar 9, 2015

### AlephNumbers

You are right. Your solution above looks good. I think that you either made a calculation mistake, or that the answer you claim is correct, is not actually correct.

19. Mar 9, 2015

### Jacky Lee

Hmm, okay. Thank you for all the help! :) I really appreciate it.

20. Mar 9, 2015

Any time.