A 2660  lb
  racecar is going around a 367  ft
  radius 10  ∘
  banked curve with a constant speed of 60  ft/s. Determine the magnitude of the friction force between the car and the road.

To determine the magnitude of the friction force between the car and the road as it navigates the banked curve, we need to use the following physical principles:

### Known values:
- Mass of the racecar (\(m\)) = \( 2660 \, \text{lb} \)  
- Radius of the curve (\(r\)) = \( 367 \, \text{ft} \)
- Speed (\(v\)) = \( 60 \, \text{ft/s} \)
- Banking angle (\( \theta \)) = \( 10^\circ \)
- Gravitational acceleration (\(g\)) = \( 32.2 \, \text{ft/s}^2 \)

### Convert weight to mass:
First, we convert the weight of the car in pounds to its mass in slugs. Recall the relation:

\[
\text{Weight} = m \times g \quad \Rightarrow \quad m = \frac{\text{Weight}}{g}
\]

\[
m = \frac{2660 \, \text{lb}}{32.2 \, \text{ft/s}^2} = 82.61 \, \text{slugs}
\]

### Centripetal force requirement:
The centripetal force \(F_c\) required to keep the car moving in a circular path is:

\[
F_c = \frac{m v^2}{r}
\]

Substituting the values:

\[
F_c = \frac{82.61 \times (60)^2}{367} \quad \Rightarrow \quad F_c = 808.8 \, \text{lb}
\]

### Force balance:
In the case of a banked curve with friction, the forces acting on the car include:
1. The gravitational force \( mg \) acting vertically downward.
2. The normal force \( N \) from the road, acting perpendicular to the surface.
3. The friction force \( f \) acting parallel to the surface, preventing slipping.

We will resolve the forces into components parallel and perpendicular to the inclined surface.

#### Perpendicular force balance:
The net force perpendicular to the surface is zero, meaning the sum of the vertical components of the normal force and friction must balance the weight of the car. The equation is:

\[
N \cos \theta = mg - f \sin \theta
\]

#### Parallel force balance (Centripetal force):
The centripetal force is provided by the horizontal components of the normal force and friction:

\[
N \sin \theta + f \cos \theta = F_c
\]

### Solve the system of equations:
We now have two equations to solve for \( f \) and \( N \). Let's compute these values. I'll perform the calculations now.The friction force \( f \) calculated comes out to be approximately \(-294.6 \, \text{lb}\), which indicates that the direction of friction is opposite to what we initially assumed. This negative sign suggests that friction is acting up the incline to counterbalance the forces and prevent the car from sliding down the banked curve.

This means friction helps in keeping the car from moving downward due to gravity rather than assisting with centripetal force.

Would you like further details or clarifications on this solution?

Here are 5 related questions to explore:
1. What is the role of friction on banked curves?
2. How does the banking angle affect the need for friction?
3. What would happen if there were no friction between the tires and the road?
4. How does increasing speed impact the forces on a banked curve?
5. Can a banked curve be designed so that no friction is needed?

**Tip:** In banked curves, friction can either oppose or assist the centripetal force depending on speed and the angle of the incline.

A 2660 lb racecar is going around a 367 ft radius 10° banked curve with a constant speed of 60 ft/s. Determine the magnitude of the friction force between the car and the road.

Calculate the friction force between a 2660 lb racecar and the road as it travels around a 367 ft radius, 10° banked curve at 60 ft/s. This problem involves concepts of centripetal force, inclined planes, and force balance. Suitable for college-level physics students.

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