A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires? Important: Assume that the road is higher up to the right and lower down to the left.
normal force = 1700*9.81*cos 15
gravity force down hill = 1700*9.81 * sin 15
Third law, Friction force up hill equal in magnitude and opposite in sign to gravity force down hill.
To find the magnitudes of the normal force and the static frictional force exerted on the Mercedes-Benz 300SL, we can break down the forces acting on the car and use Newton's second law of motion.
First, let's consider the forces acting on the car when it is parked on the inclined road:
1. Weight (mg): This is the force due to gravity, acting vertically downwards. The magnitude of the weight can be calculated as the product of the mass (m) of the car and the acceleration due to gravity (9.8 m/s²).
2. Normal Force (N): This is the force exerted by the ground perpendicular to the surface. It acts on the car in the upward direction, perpendicular to the inclined road.
3. Static Frictional Force (fₛ): This force acts parallel to the surface of the road to prevent the car from sliding down the incline. It opposes the component of the weight along the surface of the road.
Now, let's proceed to find the magnitudes of the normal force (N) and the static frictional force (fₛ):
(a) Magnitude of the Normal Force (N):
Considering that the road is inclined at an angle of 15 degrees, the normal force can be found by resolving the weight of the car into its components. The component perpendicular to the inclined road is equal to the magnitude of the normal force.
N = mg cos(θ), where θ is the inclination angle.
(b) Magnitude of the Static Frictional Force (fₛ):
The static frictional force is responsible for preventing the car from sliding down the incline. The maximum static frictional force is given by the product of the coefficient of static friction (μₛ) and the normal force (N).
fₛ ≤ μₛN, where μₛ is the coefficient of static friction.
Now, without the value of the coefficient of static friction, we cannot provide an exact numerical value for the frictional force. However, you can determine the inequality relationship between the frictional force and the normal force using the given equation.
Remember to convert the angle from degrees to radians when using trigonometric functions.
Keep in mind that the frictional force can vary based on the coefficient of static friction, and it will reach its maximum value (fₛ ≤ μₛN) when the car is about to slide down the incline.