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The normal force is a critical component in understanding how objects interact with surfaces. It is the force exerted by a surface to support the weight of an object resting on it. This force acts perpendicular to the surface. In the case of the parked Mercedes-Benz 300SL on a slope, the normal force ensures the car does not sink into the road.

To calculate the normal force (\(F_N\)) in our scenario, we first need to consider the components of gravitational force. The gravitational force pulls the car directly downward towards the earth. However, because the car is on a slope, this force has two components: one acting perpendicular to the incline and another parallel to it. The perpendicular component of the gravitational force is balanced by the normal force.

Using the angle of the incline (\(\theta = 15^\circ\)) and the cosine function, we resolve the component of gravitational force acting perpendicular to the incline:

• \(F_{\text{g}_{\perp}} = F_g \times \cos(\theta)\)
• \(F_{\text{g}_{\perp}} = 16660 \text{ N} \times 0.9659 = 16094 \text{ N}\)

Hence, the normal force on the car is equal in magnitude to \(16094 \text{ N}\), counterbalancing the perpendicular gravitational force.

Static frictional force comes into play when an object is at rest on a surface and a force attempts to move it. It is the force that prevents motion when the object is stationary. For our Mercedes-Benz parked on an incline, the static frictional force acts parallel to the surface, effectively opposing the gravitational pull that attempts to slide the car down the slope.

The magnitude of the static frictional force can be calculated once we determine the parallel component of gravitational force. This component tries to drive the car down the incline:

• \(F_{\text{g}_{\parallel}} = F_g \times \sin(\theta)\)
• \(F_{\text{g}_{\parallel}} = 16660 \text{ N} \times 0.2588 = 4310 \text{ N}\)

This force is exactly balanced by the static frictional force since the car does not move. Thus,

• \(F_s = 4310 \text{ N}\)

In essence, static friction is crucial for keeping the car parked securely on the incline, preventing any sliding.

Gravitational force can be thought of as the pull exerted by the Earth on the car, influencing everything from its movement to its state of rest on a slope. This force acts downward, but when an object is on an incline, it must be broken down into two components: perpendicular and parallel to the surface of the incline.

To understand these components:

• The **perpendicular component** (\(F_{g\perp}\)) acts at a right angle to the surface and is responsible for the normal force. This prevents the car from sinking into the road.
• The **parallel component** (\(F_{g\parallel}\)) runs along the incline, trying to pull the car down the slope, and is counteracted by the static frictional force.

The method to resolve the gravitational force includes using trigonometric functions:

• \(F_g = 16660 \text{ N}\) is the total gravitational force.
• \(F_{g\perp} = F_g \times \cos(15^\circ) = 16094 \text{ N}\)
• \(F_{g\parallel} = F_g \times \sin(15^\circ) = 4310 \text{ N}\)

These components help us understand how gravitational pull affects the car and enables us to determine the necessary opposing forces, keeping everything in balance.