Question

A curve of radius 151 m is banked at an angle of 11°. An 838-kg car negotiates the curve at 86 km/h without skidding. Neglect the effects of air drag and rolling friction. Find the following.
(a) the normal force exerted by the pavement on the tires
kN ?

(b) the frictional force exerted by the pavement on the tires
kN ?

(c) the minimum coefficient of static friction between the pavement and the tires?

Answer 1

a) 8.52 kN

b) 0.74 kN

c) 0.087

Explanation:

a)

There are 3 forces acting on the car on the banked curve:

- The weight of the car, $mg$, vertically downward

- The normal force of the pavement on the tires, N, upward perpendicular to the road

- The force of friction, $F_f$, down along the road

Resolving the 3 forces along two perpendicular directions (horizontal and vertical), we obtain the equations of motions:

x- direction:

$N sin \theta +F_f cos \theta = m \frac{v^2}{r}$ (1)

y- direction:

$Ncos \theta -F_f sin \theta-mg =0$ (2)

where

$\theta=11^{\circ}$ is the angle of the ramp

$F_f$ is the force of friction

m = 838 kg is the mass of the car

r = 151 m is the radius of the curve

$v=86 km/h =23.9 m/s$ is the speed of the car

Solving eq.(2) for Ff and substituting into eq.(1), we can find the normal force:

From (2):

$F_f=\frac{Ncos \theta-mg}{sin \theta}$ (3)

Substituting into (1) and re-arranging,

$N sin \theta +\frac{N cos \theta -mg}{sin \theta} cos \theta = m \frac{v^2}{r}\\N(sin \theta+\frac{cos^2 \theta}{sin \theta})=m\frac{v^2}{r}+mg\frac{cos \theta}{sin \theta}\\N=\frac{\frac{mv^2}{r}+mg \frac{cos \theta}{sin \theta}}{sin \theta + \frac{cos^2\theta}{sin \theta}}=8518.3 N = 8.52 kN$

b)

The frictional force between the pavement and the tires, $F_f$, can be found by using eq.(3) derived in part a):

$F_f=\frac{Ncos \theta-mg}{sin \theta}$

where we have:

$N=8518.3 N$ is the normal force

$\theta=11^{\circ}$ is the angle of the ramp

m = 838 kg is the mass of the car

$g=9.81 m/s^2$ is the acceleration due to gravity

Substituting the values, we find:

$F_f=\frac{(8518.3)cos 11^{\circ}-(838)(9.81)}{sin 11^{\circ}}=739.0 N = 0.74 kN$

c)

The force of friction between the road and the tires can be rewritten as

$F_f=\mu_s N$

where

$\mu_s$ is the coefficient of static friction

N is the normal force exerted by the road on the car

In this problem, we know that

N = 8.52 kN is the normal force

$F_f=0.74 kN$ is the frictional force

Therefore, the minimum coefficient of static friction between the pavement and the tires is:

$\mu_s = \frac{F_f}{N}=\frac{0.74}{8.52}=0.087$