A car moving at 37 km/h
 negotiates a 140 m
 -radius banked turn designed for 60 km/h
 .
What coefficient of friction is needed to keep the car on the road?

In this problem, we need to determine the coefficient of friction (\( \mu \)) required for a car moving at \( 37 \, \text{km/h} \) to stay on the road while negotiating a turn with a radius of \( 140 \, \text{m} \). The turn is banked for a speed of \( 60 \, \text{km/h} \).

### Step-by-Step Solution:

1. **Convert Speeds to SI Units (m/s)**:
   \[
   v = 37 \, \text{km/h} = \frac{37 \times 1000}{3600} \, \text{m/s} \approx 10.28 \, \text{m/s}
   \]
   \[
   v_{\text{banked}} = 60 \, \text{km/h} = \frac{60 \times 1000}{3600} \, \text{m/s} = 16.67 \, \text{m/s}
   \]

2. **Determine the banking angle**:
   The banking angle (\( \theta \)) can be calculated using the formula for a banked turn without friction at the design speed:
   \[
   \tan(\theta) = \frac{v_{\text{banked}}^2}{r g}
   \]
   where:
   - \( v_{\text{banked}} = 16.67 \, \text{m/s} \)
   - \( r = 140 \, \text{m} \)
   - \( g = 9.8 \, \text{m/s}^2 \)

   Substituting values:
   \[
   \tan(\theta) = \frac{(16.67)^2}{140 \times 9.8} = \frac{278.0889}{1372} \approx 0.2026
   \]
   \[
   \theta \approx \tan^{-1}(0.2026) \approx 11.5^\circ
   \]

3. **Apply Newton's Second Law for the car on a banked curve**:
   In a banked curve with friction, the forces in the vertical and horizontal directions are considered:
   - Horizontal force: \( f_r = m a_r = \frac{m v^2}{r} \)
   - Vertical forces include gravitational force and normal force, and the friction force provides an additional component.

   The total force in the horizontal direction (centripetal force) is:
   \[
   f_{\text{horizontal}} = N \sin(\theta) + \mu N \cos(\theta)
   \]
   and the vertical force balance (no vertical acceleration) is:
   \[
   N \cos(\theta) = mg + \mu N \sin(\theta)
   \]

4. **Solve for the Coefficient of Friction**:
   Using these two force equations, the coefficient of friction required to maintain the car on the turn can be found using:
   \[
   \mu = \frac{\frac{v^2}{r} - g \sin(\theta)}{g \cos(\theta) + \frac{v^2}{r} \sin(\theta)}
   \]
   Plugging in the values:
   - \( v = 10.28 \, \text{m/s} \)
   - \( r = 140 \, \text{m} \)
   - \( g = 9.8 \, \text{m/s}^2 \)
   - \( \theta = 11.5^\circ \)

   Let's compute the required coefficient of friction.The calculated coefficient of friction is approximately \(-0.123\). Since a negative coefficient of friction is physically unrealistic, this implies that no friction is required to keep the car on the road. In fact, friction would act in the opposite direction to prevent the car from sliding up the banked turn. This suggests that at the given speed of \(37 \, \text{km/h}\), the banking angle alone is sufficient to provide the necessary centripetal force, and the car does not need friction to stay on the road.

### Would you like more details or have any questions?

Here are five related questions to explore further:
1. What happens if the speed of the car exceeds \(60 \, \text{km/h}\)?
2. How would the required friction change if the radius of the turn were larger?
3. What is the role of friction in non-banked turns?
4. How do different road conditions (e.g., wet or icy surfaces) affect the coefficient of friction?
5. How does the angle of banking influence the speed for which a turn is designed?

**Tip**: On dry roads, a higher banking angle reduces the need for friction, making turns safer at higher speeds.

A car moving at 37 km/h negotiates a 140 m-radius banked turn designed for 60 km/h. What coefficient of friction is needed to keep the car on the road?

Learn how to calculate the coefficient of friction required for a car moving at 37 km/h on a 140m-radius banked turn designed for 60 km/h. This problem involves kinematics, dynamics, and centripetal force, using key formulas like Newton's Second Law and friction calculations. Suitable for undergraduate or advanced high school physics students.

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