Question

A car of weight \(W=\) \(10.0 \mathrm{kN}\) makes a turn on a track that is banked at an angle of \(\theta=20.0^{\circ} .\) Inside the car, hanging from a short string tied to the rear-view mirror, is an ornament. As the car turns, the ornament swings out at an angle of \(\varphi=30.0^{\circ}\) measured from the vertical inside the car. What is the force of static friction between the car and the road?

Short answer

Answer: The force of static friction between the car and the banked road during the turn is approximately 1710.1 N.

Step-by-step solution

Identify known values

We have the following known values: 1. Weight of the car, \(W = 10.0\,\text{kN}\) 2. Angle of the banked road, \(\theta = 20.0^{\circ}\) 3. Angle of the ornament from the vertical inside the car, \(\varphi = 30.0^{\circ}\)

Determine the gravitational force acting on the car

The gravitational force acting on the car is equal to its weight which is \(10.0\,\text{kN}\). We can get the force in Newtons by multiplying by \(10^3\) since \(1\,\text{kN} = 10^3\,\text{N}\). So, the gravitational force, \(F_{g} = 10.0 \times 10^3\,\text{N} = 10^4\,\text{N}\)

Determine the tilt axis and g-force

The tilt axis of the car is in the -z direction if we choose x, y, and z as the directions. The car's acceleration is creating a g-force in the direction of the tilt axis. We decompose the gravitational force into two components: one along the tilt axis (z-direction) and the other in the vertical direction (y-direction). Let's denote the g-force in the tilt axis direction as \(F_t\). This force can be found from the given angle of the ornament, \(\varphi\): \(F_t = F_g \sin{\varphi} = 10^4 \sin{30^{\circ}} = 10^4 \times 0.5 = 5000\,\text{N}\)

Determine the normal force

The normal force, \(F_{n}\), acts vertically upward, opposite to the gravitational force. We can determine it using the gravitational force component in the vertical direction and the angle \(\theta\) of the banked road. \(F_{n} = F_g \cos{(\theta + \varphi)} = 10^4 \cos{(20^{\circ} + 30^{\circ})}\) Calculate and find the value: \(F_{n} = 10^4 \cos{50^{\circ}} \approx 6427\,\text{N}\)

Calculate horizontal forces

Since the car is not moving vertically, the horizontal forces are equal and opposite. Thus, we have the static friction force, \(F_{f}\) and the centrifugal force, \(F_c\), acting on the car. Due to the tilt axis, the centrifugal force can be represented as: \(F_c = F_t \sin{\theta}\) Now, since the static friction force balances the centrifugal force: \(F_{f} = F_c\)

Determine the force of static friction

We now calculate the force of static friction using the previously calculated g-force in the tilt axis and the angle of the banked road: \(F_{f} = F_t \sin{\theta} = 5000 \sin{20^{\circ}} \approx 1710.1\,\text{N}\) So, the force of static friction between the car and the road is approximately \(1710.1\,\text{N}\).

Key concepts

Static Friction Force

When discussing a banked road physics problem, the concept of static friction is pivotal. Imagine you're in a car, turning on a road that's sloped or banked. To avoid sliding down the slope or skidding outward due to the turn, you depend on static friction. This force acts parallel to the contact surface between the car tires and the road, opposing the lateral motion that the turn attempts to induce.

Static friction increases to a maximum limit as the tendency to slide increases, aiding the car in making a safe turn. Imagine holding a book on a tilted board; it's the static friction that prevents it from sliding off until a certain angle. Similarly, for a car on a banked road, the static frictional force ensures that the vehicle maintains its circular path around the curve without losing grip on the road surface.

When calculatin the force of static friction in this context, the goal is to counteract any centrifugal force trying to push the car outward on the bank. It's calculated as the component of the g-force in the tilt axis direction, adjusted for the angle of the banked road, ensuring the car remains in a stable turn.

Gravitational Force

Gravitational force is a fundamental concept in any physics problem, including those involving banked roads. It's essentially the force of attraction exerted by the Earth on the car, which corresponds to the car’s weight. In our case, it's denoted by the symbol F_g and has been given as 10,000 N (10 kN).

This force acts downwards towards the center of the Earth and is central in determining both the normal force and the static friction force, as these are reactions to the car's weight. When the car is on a banked curve, the gravitational force does not change, but its effect is perceived differently due to the inclined surface, leading to the decomposition of this force into components parallel and perpendicular to the road surface. Understanding this concept is crucial for appreciating how the car remains on its path and does not slip off the road.

Centripetal Force

In a banked road scenario, centripetal force is the required force that keeps the car moving in a circular path. It is not a separate force, but rather the result of the net effect of other forces, such as static friction and normal force, directing the car towards the center of the circular path.

The term 'centripetal' signifies 'center-seeking' and as the car turns, this force is essential for the change in direction. Imagine swinging a ball on a string - the tension in the string directs the ball to move in a circle, similar to how the forces on the car ensure its curved trajectory on the road. The centripetal force prevents the car from continuing in a straight line, obeying Newton’s first law of motion.

Calculating the centripetal force involves understanding the car’s mass, the radius of the turn, and the speed at which the car is traveling. It is a vital component in analyzing the motion of the car on a banked road and highlights the relationship between speed, turn radius, and the ability of the car to stay on the road without sliding.

Normal Force

Normal force is an indispensable concept in banked road physics problems, as it represents the perpendicular contact force exerted by a surface on an object. For a car on a banked road, the normal force is directed perpendicular to the road surface and is the reactive force to the component of gravitational force pushing the car against the road.

The name 'normal' in this context is derived from the geometric term meaning perpendicular. It is always directed away from the surface and balances part of the car's weight. The more sharply angled the banked road is, the more significant the normal force's role in maintaining stability. It basically acts as one of the car's main supports, pushing up against gravity and preventing the car from sinking through the road surface.

In the problem, you determine the normal force considering the angle of the banked road and the car's weight. The normal force significantly influences the static friction force since it affects how much frictional resistance can be generated between the car’s tires and the road.