Problem 32 - University Physics with Modern Physics

Question

A skier of mass 65.0 \(\mathrm{kg}\) is pulled up a snow-covered slope at constant speed by a tow rope that is parallel to the ground. The ground slopes upward at a constant angle of \(26.0^{\circ}\) above the horizontal, and you can ignore friction. (a) Draw a clearly labeled free-body diagram for the skier. (b) Calculate the tension in the tow rope.

Step-by-Step Solution

Answer

The tension in the tow rope is approximately 279.3 N.

Step 1: Understand the problem

We have a skier of mass 65.0 kg being pulled up a slope that is inclined at 26.0° to the horizontal. The skier is moving at a constant speed, suggesting no net force along the slope.

Step 2: Identify the forces acting on the skier

The forces acting on the skier include the gravitational force, the tension force in the tow rope, and the normal force exerted by the slope on the skier. Since the skier moves at constant speed, the tension force must only counteract the component of gravity that acts along the slope.

Step 3: Draw the free-body diagram

In the free-body diagram, represent the skier as a dot. Draw vectors for: - The gravitational force \( F_g = mg \) acting downward. - The tension force \( F_T \) in the direction parallel to the slope, pulling the skier up. - The normal force \( F_N \) perpendicular to the slope.

Step 4: Decompose the gravitational force

Decompose the gravitational force into components parallel and perpendicular to the slope. The parallel component is given by \( F_{g, ext{parallel}} = mg \sin(26.0°) \) and the perpendicular component is \( F_{g, ext{perpendicular}} = mg \cos(26.0°)\).

Step 5: Apply equilibrium conditions

Since the skier moves at constant speed, the net force along the slope is zero. Set the tension force equal to the parallel component of the gravitational force: \( F_T = mg \sin(26.0°) . \)

Step 6: Calculate the tension in the tow rope

Substitute the values: \( m = 65.0 \ ext{kg}, g = 9.81 \ ext{m/s}^2 \). Calculate \( F_T = 65.0 \times 9.81 \times \sin(26.0°) \approx 279.3 \ ext{N} \).

Key Concepts

Free-body Diagram

A free-body diagram is an essential tool in physics, helping us visualize the forces acting on an object. When tasked with analyzing motion, like that of our skier, we start by isolating the skier in our minds and considering all forces acting upon them. In this problem:

• We portray the skier as a simple dot to simplify our scenario.
• From this point, we draw arrows to represent the forces:
  • The gravitational force (pulling downward toward the center of the Earth).
  • The tension force in the tow rope (acting parallel to the slope, pulling upward).
  • The normal force (acting perpendicular to the slope, supporting the skier).

By doing this, we can easily see the relationships and angles between the forces, particularly how gravity can be split into two components: parallel to the slope and perpendicular to the slope. This helps us calculate the necessary components further down the line.

Newton's Laws of Motion

Newton's Laws of Motion are the guiding principles when analyzing motion. They help us understand the dynamics behind our skier on the slope.

• First Law (Inertia): This law states that an object at rest remains at rest, and an object in motion continues in motion with the same speed and direction unless acted upon by an unbalanced force. For our skier, moving at constant speed means no net force is acting along the slope, implying equilibrium in the motion.


• Second Law (F=ma): This relates an object’s mass and the forces acting upon it to the acceleration that occurs. For our skier, the tension in the rope equals the gravitational component pulling them back, ensuring constant speed and zero acceleration along the slope as per the formula: \( F_T = F_{g,\text{parallel}} \).


• Third Law (Action and Reaction): Though not directly involved here, it reminds us that the forces have pairs implying the tension force is equal and opposite to the gravitational parallel component acting on the skier.

Inclined Plane

Inclined planes introduce unique dynamics into problems like these. They make us reconsider how gravitational force acts, both directly and in terms of components.

• Inclination Angle: The angle of the slope (26.0° in this exercise) dictates how forces must be resolved. For the skier, this angle affects the gravitational force components:


• The parallel component, seen as trying to pull the skier back down the slope, is calculated using \( mg \sin(\theta) \).


• The perpendicular component, countered by the normal force, is given by \( mg \cos(\theta) \).
• Force Components: These component forces are fundamental in setting the problem into equilibrium, allowing us to solve for unknowns like the tension in the tow rope.

Purposefully ignoring friction in this exercise focuses our attention solely on these resolved forces, simplifying our calculations and emphasizing the primary physics involved.