Question

A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when the coefficient of static friction between the road and rubber is 0.06, a car at rest must not slide into the ditch and a car traveling less than 50 km/h must not skid to the outside of the curve. The acceleration of gravity is 9.81 m/s 2 . θ pham (vp7427) – HW06 - Rotation – perry – (54210) 5
At what angle should the road be banked? Answer in units of ◦ . 020 (part 2 of 2) 10.0 points

What is the minimum radius of curvature of the curve? Answer in units of m. courehero

Answer 1

Answer:

1. $3.4^{o}$

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

                                              = (50)(1000/3600) m/s

                                               = 13.89 m/s

The acceleration due to gravity, $g = 9.81 m/s^{2}$

The frictional force, f = μsN     ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

        ΣF = mg sinθ-f = 0      ...... (2)

Substitute the equation (1) in equation (2), we get

 ΣF = mgsinθ-μsN = 0

 mgsinθ-μsmgcosθ =0

 μs = sinθ/cosθ

   tanθ = μs

    θ = tan-1( μs) = tan-1(0.06) = $3.4^{o}$

(b)The vertical component of the force is

N cosθ = fsinθ+mg

 N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg     ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma  (Centripetal acceleration, $a = \frac {v^{2}}{r}$

  Nsinθ+fcosθ = $m(\frac {v^{2}}{r})$

   Nsinθ+μsNcosθ = $m(\frac {v^{2}}{r})$

N[sinθ+μs cosθ] = $m(\frac {v^{2}}{r})$     ...... (4)    

Dividing the equation (4) with equation (3),

 [sinθ+μscosθ]/[cosθ- μs sinθ] = $\frac {v^{2}}{rg}$

 cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =$\frac {v^{2}}{rg}$

[tanθ+μs]/[1-μs tanθ] = $\frac {v^{2}}{rg}$      

 From part (1), tanθ = μs

 Then the above equation becomes

 $\frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

$\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

Therefore, the minimum radius of the curvature of the curve is

               $r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g}$ 

                   = $\frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}$

                   = $\frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}$

                 = 163.3 m