Banked roads -- circular motion and gravitation

  • Thread starter MPat
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  • #1
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Homework Statement


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

Homework Equations


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

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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Answers and Replies

  • #2
686
172
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
15
1
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.
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
686
172
cosθFn - sinθFfr - mg = 0 ( there is no vertical displacement)
Solve for Fn
Fn= sinθFfr + mg/cosθ

Should be:
cosθFn - sinθFfr - mg = 0
cosθFn = sinθFfr + mg
cosθFn/cosθ = (sinθFfr + mg)/cosθ
Fn = sinθFfr/cosθ + mg/cosθ
 
  • #5
15
1
Should be:
cosθFn - sinθFfr - mg = 0
cosθFn = sinθFfr + mg
cosθFn/cosθ = (sinθFfr + mg)/cosθ
Fn = sinθFfr/cosθ + mg/cosθ

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
686
172
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
15
1
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.
Thanks for your help! Much appreciated.
 
  • #8
21,466
4,854
$$F_n\cos \theta-F_{fr}\sin \theta=mg\tag{1}$$
$$F_n\sin \theta+F_{fr}\cos \theta = m\frac{v^2}{r}\tag{2}$$
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
686
172
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

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
21,466
4,854
@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
15
1
@MPat
If you had followed the simple linear equation elimination approach I indicated in post #8, you would have immediately obtained:
$$F_n=mg\cos \theta+m\frac{v^2}{r}\sin \theta$$
$$F_{fr}=m\frac{v^2}{r}\cos \theta-mg\sin \theta$$
GOT IT!!...answer is Us = 0.22
 
  • #12
haruspex
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GOT IT!!...answer is Us = 0.22
I confirm that (0.216).
 

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