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 road is approximately 3.52 kN.

Step-by-step solution

Analyze forces acting on the car

The main forces acting on the car during the turn are: 1. The weight of the car \(W\) acting vertically downward. 2. The normal force \(N\) exerted by the road surface on the car, acting perpendicular to the road surface. 3. The static friction force \(F_s\) acting horizontally along the surface of the road, which opposes the tendency of the car to slide. We need to resolve these forces into their components in the horizontal (x direction) and vertical (y direction) planes.

Resolve the weight of the car into components

We know that the weight of the car acts vertically downwards. To find the components of the weight vector along the x and y axes, we will use the given angle \(\theta\). The component of the weight vector along the x direction is: \(W_x = W \sin \theta\) The component of the weight vector along the y direction is: \(W_y = W \cos \theta\)

Resolve the normal force into components

Now, we resolve the normal force \(N\) into its components along x and y axes using angle \(\varphi\). The angle between \(N\) and \(W_y\) is \(\varphi\), so we have: The component of the normal force vector along the x direction is: \(N_x = N \cos \varphi\) The component of the normal force vector along the y direction is: \(N_y = N \sin \varphi\)

Apply Newton's second law

Using Newton's second law, we know that the sum of the forces in both directions should be equal to zero as the car is not accelerating either vertically or horizontally: In the x direction: \(N_x - W_x + F_s = 0\) In the y direction: \(N_y + W_y = 0\)

Solve for the force of static friction

From the Newton's second law equations, we can solve for \(F_s\): \(F_s = W_x - N_x\) Substitute expressions for \(W_x\) and \(N_x\): \(F_s = W \sin \theta - N \cos \varphi\) Next, we can solve for \(N\) from the y direction equation: \(N_y = -W_y \implies N\sin\varphi=-W\cos\theta \implies N=-\frac{W\cos\theta}{\sin\varphi}\) Substitute the value of \(N\) back into the equation for \(F_s\): \(F_s = W\sin\theta + \frac{W\cos\theta}{\sin\varphi}\cos\varphi\) Finally, replace the given values of \(W\), \(\theta\) and \(\varphi\): \(F_s = 10.0 \mathrm{kN}(\sin 20.0^{\circ}+\frac{\cos 20.0^{\circ}}{\sin 30.0^{\circ}}\cos 30.0^{\circ})\) Calculate the value: \(F_s \approx 3.52 \mathrm{kN}\) Therefore, the force of static friction between the car and the road is approximately \(3.52 \mathrm{kN}\).

Key concepts

Banked Road Physics

When a car navigates a turn on a banked road, the design of the road itself plays a critical role in assisting the car to maintain its path without relying solely on friction. A banked turn is angled so that the outer part of the road is higher than the inner part. This angle, denoted as \(\theta\), creates a component of gravitational force that helps to provide the necessary centripetal force to keep the vehicle on a curved path.

From the perspective of physics, the banking minimizes the dependence on the force of static friction because a component of the normal force, which is perpendicular to the road surface, now points towards the center of the circular path. This is due to the inclined surface of the bank, which creates a situation where the normal force has both a vertical and horizontal component. The horizontal component of the normal force provides some or all of the centripetal force required for the circular motion.

It's important for students to understand that if the banking angle and the speed of the car are perfectly matched, the car could potentially navigate the turn without any friction at all. However, in the real world, static friction still plays a role, especially when conditions are less than ideal or if the car's speed is not perfectly suited to the turn.

Force Analysis

Conducting a force analysis on an object involves identifying all the forces acting on it and then representing those forces through vector decomposition. In the case of the banked turn, our main forces are the weight of the car, the normal force from the road, and the static frictional force. The weight always acts directly downward, while the normal force acts perpendicular to the surface, and the static frictional force acts parallel to the surface, opposing any sliding motion.

Vector Decomposition

In order to better understand these forces and their effects on the car, we break them down into components along the x and y axes. This allows us to apply Newton's second law in each direction separately. For the weight and normal force, you consider the banking angle \(\theta\) and the angle of the swinging ornament \(\varphi\), respectively. By resolving these forces into components, we can see how they contribute to both centripetal force — required for turning — and the balancing of forces in the vertical direction to prevent the car from gaining or losing altitude on the banked curve.

Newton's Second Law

Newton's second law of motion is central to understanding how forces interact with objects. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass: \( F = ma \) where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration. When applied to a car on a banked road, we consider the car to be in a non-accelerating state — at least vertically and tangentially, meaning that any net force in these directions is zero.

As such, using Newton's second law, we can set up equations of motion in the x and y directions, summing forces to zero, since the car is neither accelerating upwards/downwards nor moving inwards/outwards on the bank. This balanced state of forces allows us to find the magnitude of static friction required to complement the centripetal force provided partially by the banking of the road — ensuring the car successfully navigates the turn without slipping.