A banked circular highway curve is designed for traffice moving at 65km/h. The radius of the curve is 200 m. Traffic is moving along the highway at 40 km/h on rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

Answer

To solve the problem of determining the minimum coefficient of friction required for cars to safely navigate a banked circular highway curve, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Designed speed (V_max) = 65 km/h
- Actual speed (V) = 40 km/h
- Radius of the curve (R) = 200 m
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Convert Speeds from km/h to m/s:**
- To convert km/h to m/s, use the conversion factor \( \frac{5}{18} \).
- \( V_{max} = 65 \times \frac{5}{18} = \frac{325}{18} \approx 18.06 \, \text{m/s} \)
- \( V = 40 \times \frac{5}{18} = \frac{200}{18} \approx 11.11 \, \text{m/s} \)

3. **Use the Formula for Minimum Coefficient of Friction:**
- The formula for the minimum coefficient of friction (μ) required for safe turning on a banked curve is given by:
\[ \mu_{min} = \frac{V^2}{Rg} \]
- Here, we will use the designed speed (V_max) for our calculations to ensure safety at maximum conditions.

4. **Substitute Values into the Formula:**
- Substitute \( V_{max} \), \( R \), and \( g \) into the formula:
\[ \mu_{min} = \frac{(18.06)^2}{200 \times 9.8} \]

5. **Calculate the Values:**
- Calculate \( (18.06)^2 \):
\[ (18.06)^2 \approx 326.16 \]
- Calculate \( 200 \times 9.8 \):
\[ 200 \times 9.8 = 1960 \]
- Now, substitute these values into the formula:
\[ \mu_{min} = \frac{326.16}{1960} \approx 0.166 \]

6. **Conclusion:**
- The minimum coefficient of friction required for cars to take the turn without sliding off the road is approximately \( \mu_{min} \approx 0.166 \).