Question

A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)

Answer 1

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

$\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625$

Weight transfer is given as follows;

$W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N$

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

$F_R$ × 1.3 = 1.7 × $F_F$

$F_R$ + $F_F$ = 18000

$F_R + \dfrac{1.3 }{1.7} \times F_R = 18000$

$F_R$ = 18000*17/30 = 10200 N

$F_F$ = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel $F_R$  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel $F_F$ =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

$\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625$

The least coefficient of friction, μ = 0.625.