Final answer:
To calculate the minimum coefficient of static friction that allows a car to make a circular turn without sliding, we use the centripetal force equation and equate it to the frictional force. After cancelling the mass, we solve for the coefficient of static friction using the velocity, gravitational acceleration, and radius of the turn.
Step-by-step explanation:
To determine the minimum oefficient of static friction that will allow a car to make a circular turn without sliding, we can use the formula for centripetal force: \( F_c = m \cdot v^2 / r \), where \( m \) is the mass of the car, \( v \) is the velocity, and \( r \) is the radius of the turn. This force is provided by the frictional force, which is the product of the coefficient of static friction (\( \mu_s \)) and the normal force (\( N \)). In this case, the normal force is equal to the gravitational force on the car, so \( N = m \cdot g \). The formula for the frictional force is thus \( f = \mu_s \cdot N = \mu_s \cdot m \cdot g \).
Since \( f = F_c \), we can equate the two expressions to find the coefficient of static friction:
\[ \mu_s \cdot m \cdot g = m \cdot v^2 / r \]
Cancelling \( m \) from both sides, we get:c
\[ \mu_s = v^2 / (g \cdot r) \]
Using the given values, \( v = 11.2 \text{m/s} \), \( g = 9.8 \text{m/s}^2 \), and \( r = 51.0 \text{m} \), the minimum coefficient of static friction can be calculated. This will provide the value of static friction needed for the car to navigate the turn without slipping.