What is the Needed Coefficient of Static Friction for a Car on a Banked Curve?

In summary: I am assuming that you know how to express the normal force in terms of the bank angle.Then find the angle for the second scenario and use that to find the coefficient of static friction. In summary, to find the coefficient of static friction for a car not to skid when traveling at 90 km/h on a curve with a radius of 80m and a bank angle of 25.8 degrees, the equations of Newton's second law (F=ma) and force equilibrium (Fx = Fc and Fnsin(theta) + Fcos(theta) =mv^2 / r) can be used. By isolating the normal force (Fn) and substituting it into the second equation, the value of the friction force (
  • #1
y90x
47
0

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 ?https://www.physicsforums.com/attachments/216722
 
Last edited:
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  • #2
y90x said:
Fncos(theta)= mg
Are these all forces acting in the vertical direction?
 
  • #3
Orodruin said:
Are these all forces acting in the vertical direction?

Yes
 
  • #4
y90x said:
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.
y90x said:
Fncos(theta)= mg
What other force has a vertical component?
Edit: Orodruin beat me to it.
 
  • #5
haruspex said:
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
haruspex said:
Orodruin beat me to it.
For once! This is a day I will tell my grandchildren about! :wink:
 
  • #7
y90x said:
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
haruspex said:
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
haruspex said:
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
y90x said:
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
haruspex said:
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
y90x said:
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
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
haruspex said:
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 into another equation to find another and all?
 
  • #15
y90x said:
To be sure , first I find one unknown variable then plug into 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
Orodruin said:
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
Thanks guys , I was able to finally solve it :) !
 

Related to What is the Needed Coefficient of Static Friction for a Car on a Banked Curve?

1) What is the "car on a bank problem"?

The "car on a bank problem" is a physics problem that involves a car parked on a hill or incline. The car is held in place by the force of friction between the tires and the road. The problem typically asks for the minimum angle of the hill at which the car will start to slide down.

2) How do you solve the "car on a bank problem"?

To solve the "car on a bank problem", you will need to use the concepts of Newton's Laws of Motion. First, draw a free-body diagram of the car and identify all the forces acting on it. Then, use the equations F=ma and ΣF=0 to find the minimum angle at which the car will start to slide. Remember to consider the direction of the forces and to use proper units.

3) What factors affect the "car on a bank problem"?

The main factors that affect the "car on a bank problem" are the force of gravity, the force of friction, and the mass of the car. The angle of the hill and the coefficient of friction between the tires and the road can also play a role in determining the minimum angle at which the car will start to slide.

4) How does the "car on a bank problem" relate to real-life situations?

The "car on a bank problem" is a simplified version of real-life situations where objects are parked or placed on an incline. It can be used to understand the balance of forces and the factors that affect the stability of an object on a slope. It is also commonly used in engineering and design to ensure the safety and stability of structures built on hills or inclines.

5) Can the "car on a bank problem" be applied to other objects?

Yes, the principles and equations used to solve the "car on a bank problem" can be applied to other objects on an incline, such as a bicycle or a box. The main concept to remember is that the object will remain stationary as long as the forces acting on it are balanced. This problem can also be extended to more complex situations, such as a car on a bank with a curved surface or a car towing a trailer on an incline.

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