Question

A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving along the highway at 37 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

Answer 1

Answer:0.077

Explanation:

Given

banked designed for traffic moving at $58 km/h\approx 16.11 m/s$

Radius of the curve 201 m

Actual traffic velocity =$37 km/h approx 10.27 m/s$

For banking of road

$tan\theta =\frac{v^2}{rg}$

$tan\theta =\frac{16.11^2}{201\times 9.81}$

$\theta =7.49^{\circ}$

Centripetal acceleration is given by

$a=\frac{v^2}{r}$

Taking component of centripetal acceleration

along and perpendicular to surface

$a_{parallel}=\frac{v^2cos\theta }{r}$

$a_{perpendicular}=\frac{v^2sin\theta }{r}$

From FBD

$mgsin\theta -f_s=ma_{parallel}$

$f_s=mgsin\theta -ma_{parallel}$----1

where $f_s$ is frictional force

$N-mgcos\theta =ma_{perpedicular}$

$N=mgcos\theta +ma_{perpedicular}$----2

and we know coefficient of friction is given by

$\mu =\frac{f_s}{N}$

$\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}$

$\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}$

$\mu =\frac{1.2804-0.5202}{9.726+0.068}$

$\mu =0.077$