Car on a bank problem need help ?!!

  • #1
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Homework Statement



If a curve with a tedious of 80m is perfectly banked for a car traveling 70km/h, what must be the coefficient of static friction for a car not to skid when traveling at 90 km/h ?

Homework Equations



Newton’s second law
F=ma

The Attempt at a Solution


To find the angle of the bank, I set the forces in the x-axis equal (because it’s perfectly banked In the first scenario) .
Fx = Fc
(Mg/cos(theta)• sin theta)=mv^2 / r
Mass cancels out
So we’re left with
g• (sin theta / cos theta) = v^2 /r
Sin over cos equals to tan , so
gtan theta = v^2/r
Isolate theta and you get
Theta = tan^-1 (v^2/r)
Theta = than^-1 (19.5^2/80)
Theta= 25.8

I will insert a photo of my free body diagram so it won’t seem confusing and my work just in case .
To find the static friction :
We can conclude that
Fnsin(theta) + Fcos(theta) =mv^2/r
And
Fncos(theta)= mg

So we isolate fn from the second equation and plug in into the first equation
So ..
(Mg/cos(theta))•sintheta + (mu)(mg/cos theta) • cos theta = mv^2/r
Simplify it to :
gtan(theta) + (mu)gcos theta= v^2/r
Isolate mu
Mu=[(v^2/r)-gtan theta]/gcos theta
When you plug in the numbers it’ll look like this
Mu= [(25^2/80)-9.8tan(25.8)]/9.8cos(25.8)
I keep getting
Mu= 0.345
The correct answer should be 0.2275

Where did I go wrong or did I miss a step ?View attachment 216722
 
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Answers and Replies

  • #2
Orodruin
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Fncos(theta)= mg
Are these all forces acting in the vertical direction?
 
  • #3
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Are these all forces acting in the vertical direction?

Yes
 
  • #4
haruspex
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Theta = tan^-1 (v^2/r)
Theta = than^-1 (19.5^2/80)
Theta= 25.8
It is hardly ever useful to find the angle. You will only be needing trig functions of the angle, and you can get all those directly from the tangent.
Fncos(theta)= mg
What other force has a vertical component?
Edit: Orodruin beat me to it.
 
  • #5
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It is hardly ever useful to find the angle. You will only be needing trig functions of the angle, and you can get all those directly from the tangent.

What other force has a vertical component?

So do I disregard the angle ?
And oh shoot ! I forgot about the vertical component of friction
 
  • #6
Orodruin
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Orodruin beat me to it.
For once! This is a day I will tell my grandchildren about! :wink:
 
  • #7
haruspex
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So do I disregard the angle ?
No, don't disregard it, just don't bother finding its value in degrees or radians straight away. Leave it as arctan(x), for a while at least. E.g. you might find at the end that all you need is the value of cos2(θ), which is just 1/(1+x2).
In my experience, working via inverse trig then trig functions worsens accuracy.
 
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  • #8
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No, don't disregard it, just don't both finding its value in degrees or radians straight away. Leave it as arctan(x), for a while at least. E.g. you might find at the end that all you need is the value of cos2(θ), which is just 1/(1+x2).
In my experience, working via inverse trig then trig functions worsens accuracy.

Okay, Noted .
Now I’m having some difficulties, so for vertical components , it will be :
Fncos(theta) = mg+ Ffsin(theta)

And the other equation being :
Fnsin(theta) + Ffcos(theta) = mv^2/r

Do I isolate Fn and substitute It into the other equation to solve for Ff ? Or do I break down Ff into mu•Fn ?
 
  • #9
Orodruin
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No, don't disregard it, just don't both finding its value in degrees or radians straight away. Leave it as arctan(x), for a while at least. E.g. you might find at the end that all you need is the value of cos2(θ), which is just 1/(1+x2).
In my experience, working via inverse trig then trig functions worsens accuracy.
This is sage advice indeed. Most of the time, it will allow you to work exclusively with rational numbers instead of with the results of inverse trigonometric functions. The OP would do well to heed this lesson.

The same thing goes for the hyperbolic functions.
 
  • #10
haruspex
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so for vertical components , it will be :
Fncos(theta) = mg+ Ffsin(theta)

And the other equation being :
Fnsin(theta) + Ffcos(theta) = mv^2/r

Do I isolate Fn and substitute It into the other equation to solve for Ff ? Or do I break down Ff into mu•Fn ?
You have three unknowns, so you need the third equation.
 
  • #11
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You have three unknowns, so you need the third equation.

Would it be to derive the mg into vertical and horizontal components as well?
 
  • #12
haruspex
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Would it be to derive the mg into vertical and horizontal components as well?
No, your third equation is the one you mentioned, the relationship between the three unknowns: μs, FN, Ff.
 
  • #13
Orodruin
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Just to say that your current system of equations (before writing down the relation between normal and friction force) has just two unknowns. You can solve for Fn and Ff easily by a rotation. (Actually, writing it down in the rotated frame from the beginning would have solved it already.) You can then apply the friction-normal relation to find the needed coefficient. I personally would find this simpler that introducing the friction-normal relation in the equations you already have.
 
  • #14
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No, your third equation is the one you mentioned, the relationship between the three unknowns: μs, FN, Ff.

To be sure , first I find one unknown variable then plug in to another equation to find another and all?
 
  • #15
Orodruin
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To be sure , first I find one unknown variable then plug in to another equation to find another and all?
I strongly suggest doing it the way suggested in #13, i.e., write down the force equilibrium in a rotated system instead. It will save you quite some computation and use of trig identities.
 
  • #16
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I strongly suggest doing it the way suggested in #13, i.e., write down the force equilibrium in a rotated system instead. It will save you quite some computation and use of trig identities.

So I just set both equations equal to either Fn or Ff and go from there ?
 
  • #17
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Thanks guys , I was able to finally solve it :) !
 

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