# Homework Help: Centripetal Force on a banked road

1. Nov 15, 2014

### spartacus

1. The problem statement, all variables and given/known data
19 m/s speed
µ = 0.2
acceleration due to gravity is 9.8 m/s2
At what angle θ should it be banked for
maximum control of the racing car?

2. Relevant equations
sum of total forces=0?
3. The attempt at a solution
N*cosθ=mg
N*sinθ-mgµ=(mv^2)/r
mg*tanθ-mgµ=(mv^2)/r
tanθ-µ=v^2/r
θ=arc tan((v^2/r)+µ)
θ=82.25 degrees?
I feel like i did everything right, seems high though...

2. Nov 15, 2014

### Staff: Mentor

No, it's accelerating.

If you want to solve for 'maximum control', I would ignore friction. But if you do include friction, which way will it act?

3. Nov 15, 2014

### spartacus

Sorry, i meant that as a relevant equation in one dimension, since it's not accelerating vertically

4. Nov 15, 2014

### Staff: Mentor

Sure, the sum of the vertical force components must be zero.

5. Nov 15, 2014

### Jazz

Does maximum control mean with aid of friction?

Are N*sinθ and mgµ acting along the same line?

6. Nov 15, 2014

### spartacus

ya so could you help me see what i did wrong?

7. Nov 15, 2014

### spartacus

I figured N*sinθ is equal to centripetal force so it would by definition be going in the opposite direction of friction

8. Nov 15, 2014

### Staff: Mentor

For one thing, you included friction in your analysis. For another, you failed to consider the direction that friction acts.

9. Nov 15, 2014

### Staff: Mentor

In the absence of friction.

What's the direction of the centripetal acceleration? What's the direction of friction with respect to the surface on which it acts?

10. Nov 15, 2014

### spartacus

Why wouldn't i include friction?
Didnt i consider the direction being against the horizontal normal force by subtracting it in N*sinθ-mgµ=(mv^2)/r

11. Nov 15, 2014

### Staff: Mentor

Because it asks for the angle of 'maximum control'. Is this question a part of a multi-part question?

Are you assuming that friction acts horizontally?

12. Nov 15, 2014

### spartacus

I believe the question means maximum control on the road that has the coefficient of friction of .2 or else why would mu be provided
and yes

13. Nov 15, 2014

### spartacus

Please dont give up on me, i easily did the other questions from this problem set, this question shouldn't be too hard, i just dont know what i did wrong

14. Nov 15, 2014

### Staff: Mentor

Note that when you consider friction, you have a range of acceptable speeds. Friction will give you the maximum and minimum values of the speeds that can be maintained without slipping. I would interpret the angle of maximum control to be that angle where friction is not needed.

The problem could just be poorly constructed. Are you posting the complete problem, word for word?

Because this could be the first part of a multipart problem. Is it?

That is incorrect. Friction acts parallel to the surface.

15. Nov 15, 2014

### Staff: Mentor

I'm not giving up on you. :)

The first thing is to make sure we have the complete problem stated. Is this from a textbook? Or devised by the instructor?

16. Nov 15, 2014

### spartacus

It's not a multipart problem no, how would i incorporate force of friction? would it be mg sintheta-mu?

17. Nov 15, 2014

### spartacus

i have it right here let ill upload a screenshot

18. Nov 15, 2014

### spartacus

19. Nov 15, 2014

### Staff: Mentor

It's a bit perplexing as to what they mean by 'maximum control'. I would interpret it as the angle that requires no friction.

Is this from some textbook?

20. Nov 15, 2014

### spartacus

No, it's an online assignment

The way i solved it, i just used the data to find which angle it would be at that speed

21. Nov 15, 2014

### Staff: Mentor

If you're going to apply friction, realize that the maximum value is given by $\mu N$ and that it acts parallel to the surface, not horizontally. And it could act up the incline or down the incline, depending on the angle.

22. Nov 15, 2014

### spartacus

If it's parallel should i use mu*mg sintheta then because that would make the problem a lot more difficult

23. Nov 15, 2014

### Staff: Mentor

It seems like a poorly worded problem to me.

24. Nov 15, 2014

### Staff: Mentor

The normal force does not equal mg.

25. Nov 15, 2014

### Staff: Mentor

I would solve the problem assuming no friction. Perhaps they are just messing with your head to see if you know what matters and what doesn't. ;)