# Min Coeff. of Friction for Banked Highway Curve @ 40 km/h

• A_lilah
In summary: Your Name]In summary, the conversation discusses how to calculate the minimum coefficient of friction required for a car to safely navigate a banked circular highway curve. The approach involves finding the angle of the banked curve, calculating the normal force and frictional force, and using the formula for coefficient of static friction. The correct answer is approximately 0.27, and further clarification is provided for some key calculations.
A_lilah

## Homework Statement

A banked circular highway curve is designed for traffic moving at 70 km/h. The radius of the curve is 193 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?

## Homework Equations

coefficient of static friction = Fstaticmax / normal force
Fnet = ma
for circular motion: Fnet = m (v^2 / radius)

## The Attempt at a Solution

y is up and down, x is left and right (up and righ are positive)
N = normal force
fstat = static friction
@ = angle

So because the bottom of the tire is always instintaneously static, I solved for the coefficient of static friction:
1. Find the angle the road is tilted at using the speed it was designed for:

70km/h = 19.44 m/s
Fnety = 0 = Ny - mg = Ncos(@) - mg

Fnetx = m(v^2/r) = Nx = Nsin(@)
solve for N... N = (mv^2) / rsin(@)

Fnety = 0 = (mv^2) / rsin(@) * cos(@) - mg
factor out mass... plug in values...

0 = [(19.44m/s^2)^2 * cos(@)] / [193m*sin(@)] - g

g = 1.958m/s^2 * [cos(@) / sin(@)]

sin(@) / cos(@) = 1.958m/s^2 / g = tan(@)

inverse tan to get @, which = 11.299 degrees

2. Find the normal force and the frictional force on the car going 40 km/h
40 km/h = 11.11 m/s
Vertical forces: weight, Ny,
Horizontal forces: fstat, Nx

sin(11.299) = Nx / N,
N = Nx / sin(11.299)

Fnetx = 0 because the minimum amount of friction should cancel out the centripital force (Nx) so the car does not slide. fstat = fstatmax in this case

Fnetx = 0 = Nsin(11.299) - fstat
fstat = Nsin(11.299)

coefficient of static friction = fstat / N
= Nsin(11.299) / N = sin(11.299) = .196

but this is not the right answer...
thanks for the help!

Thank you for your post. Your approach to solving the problem is correct, but there are a few areas that need clarification.

Firstly, your calculation for the angle of the banked curve is correct, but it should be noted that this angle is measured from the horizontal, not from the vertical. This means that the inverse tangent should be taken of the ratio of the vertical component (g) to the horizontal component (19.44 m/s^2). This will give you a steeper angle of approximately 34 degrees.

Secondly, your calculation for the normal force is incorrect. The normal force is equal to the weight of the car, which is mg. In this case, m is the mass of the car and g is the acceleration due to gravity (9.8 m/s^2). So the normal force is simply mg.

Thirdly, your calculation for the coefficient of static friction is also incorrect. The correct formula is fstat / N = (mv^2 / r) / mg. This gives a value of approximately 0.27, which is the minimum coefficient of static friction required for the car to negotiate the curve without sliding off.

I hope this helps clarify the problem. Let me know if you have any further questions.

Based on the information provided, the minimum coefficient of friction for the cars to safely negotiate the banked curve at 40 km/h is 0.196. This calculation takes into consideration the angle of the banked curve, the speed of the cars, and the centripetal force required for circular motion. However, it is important to note that this calculation assumes ideal road conditions and does not account for external factors such as the presence of water on the road. Therefore, it is recommended to exercise caution and drive at a slower speed when conditions are not ideal. Additionally, it is important for engineers and designers to consider these factors when designing highways and curves to ensure the safety of drivers.

## 1. What is the minimum coefficient of friction required for a banked highway curve at 40 km/h?

The minimum coefficient of friction required for a banked highway curve at 40 km/h is typically around 0.15.

## 2. Why is a minimum coefficient of friction necessary for a banked highway curve at 40 km/h?

A minimum coefficient of friction is necessary to prevent vehicles from slipping or sliding off the curve due to the centripetal force acting on them. Without enough friction, the vehicles may lose control and potentially cause accidents.

## 3. How is the minimum coefficient of friction calculated for a banked highway curve at 40 km/h?

The minimum coefficient of friction is calculated using the equation μ = tanθ, where μ is the coefficient of friction, and θ is the angle of the banked curve. For a banked highway curve at 40 km/h, the angle of the bank typically ranges from 10-15 degrees.

## 4. What factors can affect the minimum coefficient of friction for a banked highway curve at 40 km/h?

The main factors that can affect the minimum coefficient of friction for a banked highway curve at 40 km/h include the angle of the bank, the speed of the vehicle, the weight and distribution of the vehicle, and the condition of the road surface.

## 5. How can the minimum coefficient of friction be increased for a banked highway curve at 40 km/h?

The minimum coefficient of friction can be increased by increasing the angle of the bank, reducing the speed of the vehicle, and improving the road surface conditions. Additionally, using tires with better traction and maintaining proper vehicle maintenance can also help increase the coefficient of friction.

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