Answer to Question #219644 in Mechanical Engineering for john Smith

Question

In 1967 Mercedes-Benz presented their new steep-bank track with a maximum bank
angle of 90 degrees to the horizontal as shown below. At that time they were testing
the “new” SEC450 model that has a track width of 1.5m and its center of gravity
lies midway between the wheels. The center of gravity is 0.75 meters above the
horizontal and the coefficient of friction between the tyres and the road surface is 0.7
under dry conditions.

Determine the following:

a) The maximum speed in km/h at which this car could round an unbanked
bend of 75 meters radius

b) The angle to the horizontal the driver must bank to ensure “no-side slip”
conditions at a speed of 60 km/h

c) If the driver moves up the incline to 15 degrees to the horizontal, determine
the speed at which the car in (a) above would now skid in this bend.

Answer 1

Part a

$\frac{mv^2}{r}= \mu mg\\ v= \sqrt{\mu r g}\\ v= \sqrt{0.7*75*9.81}\\ v= 22.7 m/s=81.72 km/hr$

Part b

$\tan \theta = \frac{v^2}{rg}\\ \tan \theta = \frac{16.67^2}{75*9.81}\\ \theta = \tan^{-1}(\frac{16.67^2}{75*9.81})\\ \theta =1.26^0$

Part c

$v^2 = rg \tan \theta \\ v^2 = 75*9.81 \tan 15 \\ v^2 = 14.04 m/s$