Can you solve this for me.?

Question

a car on a horizontal road makes an emergency stop such that all four wheels lock and slide.
the coefficient of friction is 0.40. The whells are 4.2m apart
the center of mass is 1.8m behind and 0.75m above the front axle. The car weighs 1.1x10^4 N.
Calculate acceleration
Fn and Ff on each wheel

Answer 1

The frictional force F = k*W; the acceleration will be F/m, where m = W/g, then

a = k*W/[W/g] = k*g

Rotational equilibrium requires that 1.8*F1 = (4.2-1.8)*F2, where F1 is the normal force on the two front wheels and F2 the normal force on the two rear wheels. Also, F1+F2=1.1*10^4 N; Solve for F1 and F2. The normal force on each of the front wheels is (F1)/2, and on each of the rear wheels (F2)/2. The frictional force on each wheel is its normal force times k.

( I assumed the CG is centered between the wheels, since no lateral position was specified.)

Answer 2

For the rear wheels,
1.1*10^4)(1.8) = 4.2(2F)
Fn = 2,357.143 N
Ff = 0.4*Fn = 942.8571 N

For the front wheels,
2Fn = 11,000 - 2357.143 N
Fn = 4,321.429 N
Ff = 1,728.571 N