n the figure the driver of a car on a horizontal road makes

n the figure, the driver of a car on a horizontal road makes an emergency stop by applying the brakes so that all four wheels lock and skid along the road. The coefficient of kinetic friction between ires and road is 0.38. The separation between the front and rear axles is L = 4.5 m, and the center of mass of the car is located at distance d = 2.3 m behind the front axle and distance h = 1.0 m above the road. The car weighs 9.9 kN. Find the magnitude of the braking acceleration of the car, the normal force on each rear wheel, the normal force on each front wheel, the raking force on each rear wheel, and the braking force on each front wheel.

Solution

Found this explanation on another site while I was trying to do this same problem. It uses different numbers but its the right way, it worked for me at least. Hope it helps and good luck

a ) Given that weight of the car is W = mg = 9.9kN = 9900 N
Then mass of the car is m = W/g = 9900/9.8 = 1010.204 kg
We know that breaking force F = ma = -kmg
k= 0.4
Then braking acceleration of the car is a = -kg = 0.4 *9.8 =- 3.92m/s2
b) Total frictional force on the car is f = kmg = 0.4*9900 N = 3960N
Now for the car to be in equilibrium then net force acting on the car to be zero and also net torque acting on the car to be zero.
Let normal force on the rear wheel of the car be Fnr \'
normal force on the front wheel of the car be Fnf \'
h = 1.0 m
L = 4.5 m
d = 2.3m
Then Fnr + Fnf = mg -------1
Fnr + Fnf = 9900 N -------2
Then f * h +Fnr*(L-d) -Fnf*d = 0 -----------------3
3960 N * 1.0 m + Fnr*(4.5m -2.3m) -Fnf*2.3m = 0 -----------4
Solving equations 2 and 4 we get
Fnr = 3365N
This total normal force on the rear wheel is divided into two parts because the car consists of 2 rear wheels
Then normal force on each rear wheel is Fnr \' = Fnr/2 = 1682 N
c ) Now Fnf =5935
Now normal force on each front wheel is Fnf \' = Fnf/2 = 2967.5 N
d) breaking force on each rear wheel is frear = kFnr \' = 0.4 *1682 = 672 N
e ) breaking force on each front wheel is ffront = kFnf \' = 0.4*2967.5 = 1187 N