# Static Equilibrium / Normal Forces / Braking Forces on tires

• Ghostscythe
In summary: Note that the rotation equation is taken around the centre of mass.Note that this makes the clockwise force on the front wheels have no torque.Note that the clockwise force on the rear wheels is offset by the anticlockwise force on the rear wheels, and the anticlockwise friction force, all based on the distance from CoM to the point where they are applied.Note that the anticlockwise force on the rear wheels is equal to the force on the front wheel, and the friction force, so the anticlockwise torque on the rear wheels is equal to the clockwise torque on the front wheels.Note that the anticlockwise torque on the rear wheels is 1.8 * 3825 N, so the clockwise torque on the

## 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.

$f_k = .36$
L = 4.4m
d = 1.8m
L-d = 2.6m
h = .75m
W = 11,000 N
m = 1,121.30 kg

## Homework Equations

$F_Friction = f_k * F_N F = ma \tau = {r_\bot}F \tau = rF_\bot$

## The Attempt at a Solution

a) $f_k * F_N = .36 * 11000 = 3960 N$ (Correct)
b)$\tau_{front} + \tau_{back} = \tau_{Total}$
I use the front wheel as the moment so \tau_front will be 0.
$0 + rF_{back}_\bot = 1.8*F_{N}_{Total}$
4.4F_back_\bot = 1.8(11000)
F_back_\bot = 4500[/itex]
So $F_N$ 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.

Ghostscythe said:

## 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.

$f_k = .36$
L = 4.4m
d = 1.8m
L-d = 2.6m
h = .75m
W = 11,000 N
m = 1,121.30 kg

## Homework Equations

$F_Friction = f_k * F_N F = ma \tau = {r_\bot}F \tau = rF_\bot$

## The Attempt at a Solution

a) $f_k * F_N = .36 * 11000 = 3960 N$ (Correct)
b)$\tau_{front} + \tau_{back} = \tau_{Total}$
I use the front wheel as the moment so \tau_front will be 0.
$0 + rF_{back}_\bot = 1.8*F_{N}_{Total}$
4.4F_back_\bot = 1.8(11000)
F_back_\bot = 4500[/itex]
So $F_N$ 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.

Just checking: was the Reaction force on the rear wheels given as 3825 N → 1922.5 each?

Last edited:
Yes, they're equally balanced on either side of the CoM.

Ghostscythe said:
Yes, they're equally balanced on either side of the CoM.

My real question referred to the quantity.

You had referred to your answer 4500 N [divided 2250 N each wheel] as being incorrect, ad you were not sure why it was incorrect.

I had tried a slightly different approach and got 3825 N , so divided equally between the back wheels.

I was querying the 3825 N [or was it 3845?] as the correct answer gven - as it did not seem appropriate to enter a lengthy analysis to teh wrong answer.

Ghostscythe said:
Yes, they're equally balanced on either side of the CoM.

Assuming my suggested answer was correct - I proceded as follows.

I took rotation around the centre of mass.

The Upward force on the front wheels is clockwise.

The upward force on the rear wheels, and the firction force - whether it be provided by the front or rear wheels - are both anticlockwise.

Although the equation derived above has two unknowns, we also know that the sum of the upward force on the front wheels and back wheel totals the weight of the vehicle, so simultaneous equation methods can be used to solve.

At first I only calculated the upward force on the rear wheels - but it is simple to calculate the force on the front, and then split the braking force into friction force on appropriate wheels.

Yes, your calculation for the reaction force on rear wheels was right. How did you get that? I think I understand how you found the rest, but that's not clear to me.

Ghostscythe said:
Yes, your calculation for the reaction force on rear wheels was right. How did you get that? I think I understand how you found the rest, but that's not clear to me.

Did my second post explain?

## 1. What is static equilibrium?

Static equilibrium refers to a state in which all forces acting on an object are balanced, resulting in no net acceleration or movement. This means that the object is either at rest or moving at a constant velocity.

## 2. How are normal forces related to static equilibrium?

Normal forces are perpendicular forces exerted by a surface on an object in contact with it. In the case of static equilibrium, the normal force must be equal and opposite to the force exerted by the object on the surface in order for the object to remain at rest or in constant motion.

## 3. What is the role of braking forces on tires in static equilibrium?

Braking forces on tires are responsible for slowing down or stopping the motion of a vehicle. In the case of static equilibrium, the braking force must be balanced by the normal force in order to prevent the tires from slipping or skidding.

## 4. How do you calculate normal forces in static equilibrium?

In order to calculate normal forces in static equilibrium, you must first identify all the forces acting on the object and draw a free body diagram. Then, using the principle of equilibrium, you can set up equations to determine the magnitude and direction of the normal forces.

## 5. What factors can affect the normal forces in static equilibrium?

The normal forces in static equilibrium can be affected by the mass of the object, the angle of the surface, and any external forces acting on the object. Additionally, the coefficient of friction between the object and the surface can also impact the normal forces.