Problem 9 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A skier moves down a \(27^{\circ}\) slope at constant speed. What can you say about the coefficient of friction, \(\mu_{\mathrm{k}}\) ? Assume the speed is low enough that air resistance can be ignored.

Step-by-Step Solution

Answer

The coefficient of friction \( \mu_{\mathrm{k}} \) is approximately 0.51.

Step 1: Understanding Forces on the Slope

When a skier moves at constant speed down a slope, the net force acting along the direction of the slope must be zero. This is because the skier does not accelerate. The forces involved are the gravitational force component along the slope and the frictional force.

Step 2: Resolve Gravitational Force

The gravitational force acting on the skier can be broken down into two components: one perpendicular to the slope and one parallel to the slope. The parallel component, which causes the skier to slide down, is given by: \[ F_{ ext{gravity, parallel}} = mg \sin(\theta) \]where \( m \) is the mass of the skier, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the slope (27° in this case).

Step 3: Frictional Force on the Slope

The frictional force that opposes the motion is given by:\[ F_{ ext{friction}} = \mu_{\mathrm{k}} N \]where \( \mu_{\mathrm{k}} \) is the coefficient of kinetic friction and \( N \) is the normal force. The normal force is the component of gravitational force perpendicular to the slope, calculated as:\[ N = mg \cos(\theta) \].

Step 4: Equilibrium Condition for Constant Speed

Because the skier moves at a constant speed, the net force is zero. Therefore, the gravitational force parallel to the slope is balanced by the frictional force:\[ mg \sin(\theta) = \mu_{\mathrm{k}} mg \cos(\theta) \].

Step 5: Solving for the Coefficient of Friction

Simplify the equation by dividing both sides by \( mg \) to solve for \( \mu_{\mathrm{k}} \):\[ \mu_{\mathrm{k}} = \tan(\theta) \]Substitute \( \theta = 27^{\circ} \):\[ \mu_{\mathrm{k}} = \tan(27^{\circ}) \approx 0.51 \].

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept in physics. It is the force that attracts any two objects with mass toward each other. On Earth, it gives weight to objects and pulls them toward the ground. For a skier on a slope, the gravitational force can be split into two components:

• One component acts parallel to the slope, pulling the skier downward along the slope.
• The other component acts perpendicular to the slope, pressing the skier against it.

The skier's weight, calculated as the product of mass \( m \) and the acceleration due to gravity \( g \), can be divided into these components using trigonometric functions of the slope angle \( \theta \). The parallel component is responsible for the skier’s motion down the hill and is given by: \[ F_{\text{gravity, parallel}} = mg \sin(\theta) \]Understanding this force component is key to analyzing the skier's motion as it is countered by the frictional force.

Normal Force

The normal force is another crucial concept when dealing with motion on inclined planes. It acts perpendicular to the surface of contact and prevents objects from 'falling through' the surface. For a skier on a slope, the normal force provides a balancing act to the perpendicular component of the gravitational force.
Normal force can be calculated using the perpendicular component of gravitational force:\[ N = mg \cos(\theta) \]
Here, \( N \) is the normal force, \( m \) is the mass, \( g \) is gravitational acceleration, and \( \theta \) is the angle of the slope. The normal force is significant because it directly affects the frictional force, which is in turn dependent on the normal force. In the case of our skier, despite moving down the slope, the normal force remains directed upwards perpendicular to the slope.

Frictional Force

The frictional force plays a crucial role in controlling the motion of objects on a slope. It acts opposite to the direction of motion and is calculated as the product of the coefficient of kinetic friction \( \mu_{\text{k}} \) and the normal force \( N \):\[ F_{\text{friction}} = \mu_{\text{k}} N \]
This means the frictional force is dependent on both the texture of the surfaces in contact (expressed by \( \mu_{\text{k}} \)) and how hard they are pressed against each other (expressed by \( N \)). For the skier moving at a constant speed, it indicates equilibrium between gravitational pull and frictional resistance.
The skier moving down at constant speed implies that the frictional force perfectly balances the gravitational force parallel to the slope. Hence, the coefficient of kinetic friction equals the tangent of the slope's angle, a relationship derived from: \[ \mu_{\text{k}} = \tan(\theta) \] This equation helps to understand how the slope's steepness impacts the required friction to maintain a steady descent.