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## Homework Statement

In the figure below, 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 0.36. The separation between the front and rear axles is L = 4.4 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. (Hint: Although the car is not in translational equilibrium, it is in rotational equilibrium.)

(a) The braking acceleration of the car.

(b) The normal force on each rear wheel.

(c) the normal force on each front wheel.

(d) the braking force on each rear wheel.

(e) the braking force on each front wheel.

[itex]f_k = .36[/itex]

L = 4.4m

d = 1.8m

L-d = 2.6m

h = .75m

W = 11,000 N

m = 1,121.30 kg

## Homework Equations

[itex]F_Friction = f_k * F_N

F = ma

\tau = {r_\bot}F

\tau = rF_\bot[/itex]

## The Attempt at a Solution

a) [itex]f_k * F_N = .36 * 11000 = 3960 N[/itex] (Correct)

b)[itex]\tau_{front} + \tau_{back} = \tau_{Total}[/itex]

I use the front wheel as the moment so \tau_front will be 0.

[itex]0 + rF_{back}_\bot = 1.8*F_{N}_{Total}[/itex]

4.4F_back_\bot = 1.8(11000)

F_back_\bot = 4500[/itex]

So [itex]F_N[/itex] each wheel would be 2250 N, if it was correct (it isn't). Obviously no point in doing c) if I don't have the concept down.

for d) and e) I assume you do the same kind of conservation of torque, except you use braking force instead of normal force?

Thanks. Sorry about the broken itex, too pissed atm to fix it.