- #1

A_lilah

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## 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!