. A car is safely negotiating an unbanked circular turn at a speed of 19 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one third of its dry-road value. If the car is to continue safely around the curve, to what speed must the dirver slow the car?
To find the speed at which the driver must slow the car in order to continue safely around the curve, we need to ensure that the force of static friction, which provides the centripetal force for the car's circular motion, is still sufficient.
Let's break down the problem step by step:
Step 1: Find the maximum static frictional force on the wet road.
Since the driver is already negotiating the turn safely at a speed of 19 m/s on the dry road, we can assume that the maximum static frictional force on the dry road is enough to provide the necessary centripetal force.
So, let's define:
- μdry = coefficient of static friction on dry road
- μwet = coefficient of static friction on wet road
- Fdry = maximum static frictional force on dry road
- Fwet = maximum static frictional force on wet road
Given that the maximum static frictional force on the wet road (Fwet) is one-third of its dry-road value, we can write the equation:
Fwet = (1/3) * Fdry
Step 2: Equate the maximum static frictional force to the necessary centripetal force.
The maximum static frictional force is given by F = μ * N, where N is the normal force. Since the car is safely negotiating the unbanked turn, the normal force is equal to the car's weight, which can be calculated as:
N = mg, where m is the mass of the car and g is the acceleration due to gravity.
Equate the maximum static frictional force (Fwet) on the wet road to the necessary centripetal force (Fc):
Fwet = Fc
Step 3: Calculate the necessary centripetal force.
The centripetal force needed to make the car negotiate the turn safely is given by:
Fc = (mv²) / r, where m is the mass of the car, v is the velocity, and r is the radius of the circular turn.
Step 4: Equate the necessary centripetal force to the maximum static frictional force on the wet road.
Equate the necessary centripetal force (Fc) to the maximum static frictional force on the wet road (Fwet):
Fc = Fwet
Step 5: Substitute the values and solve for the unknown.
Substituting the values into the equation, we have:
(mv²) / r = (1/3) * Fdry
Now, we can solve for the speed (v) at which the driver must slow the car.