A car weighting 10.7 kN and travelling at 13.4 m/s without negative lift attmepts to round an unbanked curve with a radius of 61.0 m. (a) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction betwen the tires and the road is 0.350, is the attempt at taking the curve successful?

Answer

To solve the problem, we need to find the required frictional force to keep the car on its circular path and determine if the attempt at taking the curve is successful based on the coefficient of static friction.

### Step 1: Calculate the mass of the car
The weight of the car is given as 10.7 kN. To find the mass (m), we can use the formula:
\[ m = \frac{W}{g} \]
where \( W \) is the weight and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

\[ m = \frac{10.7 \times 10^3 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 1091.84 \, \text{kg} \]

### Step 2: Calculate the required centripetal force
The required centripetal force (\( F_c \)) to keep the car moving in a circular path can be calculated using the formula:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the speed of the car (13.4 m/s) and \( r \) is the radius of the curve (61.0 m).

\[ F_c = \frac{1091.84 \, \text{kg} \times (13.4 \, \text{m/s})^2}{61.0 \, \text{m}} \]

Calculating this gives:
\[ F_c = \frac{1091.84 \times 179.56}{61.0} \approx 3,217.58 \, \text{N} \]

### Step 3: Determine the maximum static friction force
The maximum static friction force (\( F_{\text{max}} \)) can be calculated using the formula:
\[ F_{\text{max}} = \mu_s \times N \]
where \( \mu_s \) is the coefficient of static friction (0.350) and \( N \) is the normal force, which is equal to the weight of the car.

\[ F_{\text{max}} = 0.350 \times 10.7 \times 10^3 \, \text{N} \approx 3,745 \, \text{N} \]

### Step 4: Compare the required centripetal force with the maximum static friction force
Now we compare the required centripetal force with the maximum static friction force:
- Required centripetal force: \( 3,217.58 \, \text{N} \)
- Maximum static friction force: \( 3,745 \, \text{N} \)

Since \( 3,217.58 \, \text{N} < 3,745 \, \text{N} \), the frictional force required to keep the car on its circular path is less than the maximum static friction force available.

### Conclusion
(a) The magnitude of the frictional force required to keep the car on its circular path is approximately **3,217.58 N**.

(b) Since the required frictional force is less than the maximum static friction force, the attempt at taking the curve is **successful**.

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