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 tires and road is 0.39. The separation between the front and rear axles is L = 4.1 m, and the center of mass of the car is located at distance d = 1.8 m behind the front axle and distance h = 0.75 m above the road. The car weighs 11 kN. Find the magnitude of the following

I have figured out:
The braking acceleration of the car: 3.822

and looking for:

The normal force on each rear wheel.
N

(c) the normal force on each front wheel.
N

(d) the braking force on each rear wheel.
N

(e) the braking force on each front wheel.
N

I cant quite figure THIS out i know i have to solve through simultaneous equations for F front but i have some mathematical problem ...

Please help !

shut up

The normal force on each rear wheel.

92N

(c) the normal force on each front wheel.
88N

(d) the braking force on each rear wheel.
23N

(e) the braking force on each front wheel.
67N

To solve this problem, we can start by analyzing the forces acting on the car.

When the car is braking, the only horizontal force acting on it is the force of friction between the tires and the road. Therefore, the net force on the car is equal to the force of friction.

(a) The net force on the car is given by the equation:
Net force = mass of the car * acceleration

The weight of the car is 11 kN, which means the mass is 11000 N / 9.8 m/s^2 = 1122.45 kg.

Since the net force and acceleration are in the horizontal direction, we can write:
Net force = Force of friction = Friction coefficient * Normal force

The acceleration is the braking acceleration, which is 3.822 m/s^2.

Substituting these values into the equation, we get:
1122.45 kg * 3.822 m/s^2 = 0.39 * (Normal force on each rear wheel + Normal force on each front wheel)

(b) The normal force on each rear wheel and front wheel can be found by considering the torque acting on the car.

Torque = Force * Distance

The torque due to the weight of the car acts at the center of mass, which is located 1.8 m behind the front axle. The torque due to the normal force on each wheel acts at the separation L = 4.1 m.

Setting up the torque equation around the front axle:
Torque rear wheels (due to weight) + Torque front wheels = 0

The torque due to the weight of the car is given by:
Torque rear wheels (due to weight) = weight * distance from center of mass to rear axle
Torque rear wheels (due to weight) = 11000 N * 1.8 m

The torque due to the normal force on each wheel is given by:
Torque front wheels = (Normal force on each front wheel) * L

Setting up the torque equation, we get:
11000 N * 1.8 m + (Normal force on each front wheel) * 4.1 m = 0

Solving this equation will give you the value of the normal force on each front wheel.

Once you have the value of the normal force on each front wheel, you can substitute it back into equation (a) to find the value of the normal force on each rear wheel.

(c) The normal force on each front wheel is the value you found by solving the torque equation.

(d) The braking force on each rear wheel can be found using the friction equation:
Friction force = Friction coefficient * Normal force

Substituting the known values, we get:
Friction force = 0.39 * (Normal force on each rear wheel)

(e) The braking force on each front wheel can be found using the same friction equation:
Friction force = Friction coefficient * Normal force

Substituting the known values, we get:
Friction force = 0.39 * (Normal force on each front wheel)

By solving equation (a), the torque equation, and using the friction equation, you should be able to calculate all the required values.