Banked Curve - Minimum Turn Radius

AI Thread Summary
A car traveling at 10 m/s on a banked track with a 5-degree angle and a friction coefficient of 0.3 requires a minimum radius to avoid slipping. The initial calculations suggest a radius of 34.1 m, but the book states the answer is 26 m, leading to confusion about the derivation of the forces involved. Key forces include the centripetal force, normal force, and friction, with the latter preventing slipping towards the center of curvature. Misunderstandings arise regarding the decomposition of forces and the role of the bank angle, which is not a force itself. Clarification on these concepts is essential for solving the problem accurately.
Devtycoon
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Homework Statement


A car traveling 10 m/s is moving along a track banked at 5 degrees. The tire-road friction coefficient is .3 What is the minimum radius it can travel without slipping?

v0= 10 m/s
Bank Angle Θ = 5°
μ=0.3

Note I am working through prep material for the exam. The solution in the book skips steps and doesn't make sense.

Homework Equations


Normal and Tangential Kinetics for Planar Problems - FE Exam Reference Guide Page 70 - Dynamics
∑Ft=mat
∑Fn=mvt/ρ , where ρ is the radius of curvature

The Attempt at a Solution


∑Fx=Fc-Fμ
∑Fx=mv2/ρ-μmgcosΘ=0, where Θ=5°
ρ=v2/μgcosΘ=100/.3⋅*9.81⋅cos(5°)=34.1 mNote:

Why is the review manual saying to use the following solution? I do not understand the FΘ term or how it was derived.

Fc=Fμ+Fθ
mv2/ρ=μmg+mg⋅tanΘ
ρ=26m
Snapshot.jpg
Note: Thanks for the help. I don't understand the solution and my attempt is the wrong answer.
 
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##F_c = F_\mu + F_\theta## appears to decompose the centripetal force into a part caused by friction with coefficient ##\mu## and a part caused by the inward radial component of the normal force exerted by banked track.

So you know that the banked track is pushing normal to its force with some force ##F## (acting at angle ##\theta##relative to vertical). The vertical component of ##F## balances downward gravity, so
$$F \cos\theta = mg.$$
Meanwhile, the inward radial component of ##F## is what's left, namely
$$F_\theta = F\sin\theta.$$
Now combine them to get an expression for ##F_\theta## as a function of ##mg## and ##\theta##.

Also, I wonder if you are supposed to use the small angle approxation to get ##F_\mu = \mu mg##.
 
I am confused. Is my FBD messed up? I see a gravitational force in the direction of the centripetal force.

Snapshot.jpg
 
Devtycoon said:
I am confused. Is my FBD messed up? I see a gravitational force in the direction of the centripetal force.

View attachment 87216
The centripetal force must be horizontal, as the car moves along a horizontal circle.
 
Car is on a 5 degree bank. Did I still mess up the FBD?

Snapshot.jpg
 
There is a normal force FN which is perpendicular to the road and static friction Fs that is parallel with it. The force of friction acts inward in case of smallest possible radius, to prevent the car sliding out.
bankedfr.JPG
 
I am confused because the solutions states:

Fc = Ff + FθThe overhead view looks like this. Right?
Snapshot.jpg
 
I do not understand your notations.
The friction is not horizontal.
Decompose all forces into horizontal and vertical components.
 
First of all. Thanks for your help. I clearly am having a physics fundamental issue here. I have a fundamental question before I rehash the scenario here.Centripetal force is sucking the car in towards the instantaneous center of rotation right?

Snapshot.jpg
 
  • #10
The centripetal force points towards the centre of curvature. But it does not "suck the car". It is not an operating force: it is the resultant of all forces acting on the car.
What do you mean on IoCp?
 
  • #11
ehild said:
The centripetal force points towards the centre of curvature. But it does not "sucks the car". It is not an operating force: it is the resultant of all forces acting on the car.
What do you mean on IoCp?

I got sloppy with the paint tool. Sorry. It was supposed to be I.C. , the acronym for instantaneous center.

Okay next fundamental question, there are 3 additional forces acting on the particle. The friction force, force due to bank angle and the normal force due to gravity.

1) Is the force of friction preventing the car from slipping towards or away from the center of curvature?

2) What does the book mean when it states "The bank angle is the centrifugal force contributing to the normal force."
 
  • #12
Devtycoon said:
Okay next fundamental question, there are 3 additional forces acting on the particle. The friction force, force due to bank angle and the normal force due to gravity.

There is no force due to bank angle, and the normal force is not due to gravity.
The forces are gravity, normal force (due to the road) and friction.

Devtycoon said:
1) Is the force of friction preventing the car from slipping towards or away from the center of curvature?

You have to figure it out. The force of static friction is less or equal to μN.

Devtycoon said:
2) What does the book mean when it states "The bank angle is the centrifugal force contributing to the normal force."

That statement has no sense. The bank angle is not a force.
 
  • #13
How can I thank you for your help?
 
  • #14
Devtycoon said:
How can I thank you for your help?
Show your (correct) solution :)
 
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