# Banked roads -- circular motion and gravitation

1. Aug 12, 2016

### MPat

1. The problem statement, all variables and given/known data
If a curve with a radius of 88m is perfectly banked for a car traveling at 75km/h, what must be the coefficient of static friction for a car not to skid when traveling at 95 km/hr

75km/h = 20.8m/s
95km/h = 26.4m/s

2. Relevant equations
Fr=mv^2/r
Fr=centripetal force
m = mass
v = velocity
Fn= normal force
Ffr= force of friction (Us*Fn)
g = gravitational acceleration
Us = coefficient of static friction

3. The attempt at a solution
Horizontal component
sinθFn=mv^2/r Eq 1

Vertical component
cosθFn=mg
Fn=mg/cosθ Eq 2

Sub eq 2 into 1

sinθ (mg/cosθ) = mv^2/r (sinθ/cosθ = tanθ)
tanθ mg = mv^2/r
m cancels out
tanθ = v^2/gr
θ = arctan 20.8^2/9.8*88
θ = 26.6 degrees

Am I right up to there?

Now for the second part of the question. Solving for Us.

See diagram for force of friction and horizontal and vertical components
Vertical component
cosθFn - sinθFfr - mg = 0 ( there is no vertical displacement)
Solve for Fn
Fn= (sinθFfr + mg)/cosθ
Horizontal component
sinθFn + cosθFfr = mv^2/r
Sub Fn from above

sinθ ((sinθFfr + mg)/cosθ) + cosθFfr = mv^2/r

From here I am unable to isolate Ffr, or cancel out the mass.... Help!!

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Last edited: Aug 12, 2016
2. Aug 12, 2016

### TomHart

Without going through your whole solution, one thing I noticed is that when you solve for Fn, you divided mg by cosθ, but you forgot to divide the first term by cosθ.

EDIT: Also, I don't know why you are not able to isolate Ffr.

3. Aug 12, 2016

### MPat

Under vertical component?

Vertical component
cosθFn=mg
Fn=mg/cosθ Eq 2

I divide both sides by cosθ here on the left side cosθ cancels out and on the right it becomes mg/cosθ

Is this what you were referring to?

4. Aug 12, 2016

### TomHart

Should be:
cosθFn - sinθFfr - mg = 0
cosθFn = sinθFfr + mg
cosθFn/cosθ = (sinθFfr + mg)/cosθ
Fn = sinθFfr/cosθ + mg/cosθ

5. Aug 12, 2016

### MPat

Ok that correcting that I get the following
sinθ (sinθFfr/cosθ + mg/cosθ) + cosθFfr = mv^2/r
=sinθtanθFfr + tanθmg + cosθFfr = mv^2/r
= Ffr(sinθtanθ + cosθ) + tanθmg = mv^2/r
Ffr = (mv^2/r - tanθmg)/(sinθtanθ+cosθ)

Even if I isolate Ffr here, m is unknown and I haven't been able to cancel it out...

6. Aug 12, 2016

### TomHart

MPat, I see your point. And I wish I had time right now to sort this out, but I have to leave right now. I worked out the problem and, as expected, the mass does cancel out at some point. But I don't have time to compare my solution with yours. I will try to look at it later on tonight. But hopefully, someone will beat me to it.

7. Aug 12, 2016

### MPat

Thanks for your help! Much appreciated.

8. Aug 12, 2016

### Staff: Mentor

$$F_n\cos \theta-F_{fr}\sin \theta=mg\tag{1}$$
$$F_n\sin \theta+F_{fr}\cos \theta = m\frac{v^2}{r}\tag{2}$$
Multiply Eqn. 1 by $\cos \theta$ and Eqn. 2 by $\sin \theta$. Then add them together. What do you get?

Multiply Eqn. 2 by $\cos \theta$ and Eqn. 1 by $\sin \theta$. Then subtract the result from Eqn. 1 from the result from Eqn. 2. What do you get?

9. Aug 13, 2016

### TomHart

MPat, I think your problem starts when you "solve for Fn". Mathematically, you did it right. But you really still have another Fn term on the right-hand side of that equation because the friction force Ffr is proportional to the normal force Fn. You should make that substitution BEFORE you solve for Fn.

10. Aug 15, 2016

### Staff: Mentor

@MPat
If you had followed the simple linear equation elimination approach I indicated in post #8, you would have immediately obtained:
$$F_n=mg\cos \theta+m\frac{v^2}{r}\sin \theta$$
$$F_{fr}=m\frac{v^2}{r}\cos \theta-mg\sin \theta$$

11. Aug 16, 2016

### MPat

GOT IT!!...answer is Us = 0.22

12. Aug 17, 2016

### haruspex

I confirm that (0.216).