Problem 17 - Applied Physics

Question

A tractor tows a barge through a canal with a towrope that makes an angle of \(21^{\circ}\) with the bank of the canal. If the tension in the rope is \(12,000 \mathrm{~N}\), how much work is done in moving the barge \(550 \mathrm{~m} ?\)

Step-by-Step Solution

Answer

Approximately 6,160,320 J of work is done.

Step 1: Understand the Concept of Work

The work done by a force is calculated by using the formula \( W = F \, d \, \cos(\theta) \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement.

Step 2: Identify Given Values

We know the tension in the rope is \( F = 12,000 \mathrm{~N} \), the displacement \( d = 550 \mathrm{~m} \), and the angle \( \theta = 21^{\circ} \).

Step 3: Calculate the Work Done

Using the given values in the work formula: \( W = 12,000 \cdot 550 \cdot \cos(21^{\circ}) \). Calculate \( \cos(21^{\circ}) \) which is approximately 0.9336, then calculate:\( W = 12,000 \cdot 550 \cdot 0.9336 \).

Step 4: Perform the Multiplication

Calculate \( 12,000 \cdot 550 = 6,600,000 \), and then multiply by 0.9336 to find \( 6,600,000 \cdot 0.9336 \).

Step 5: Final Calculation and Answer

After performing the multiplication \( 6,600,000 \cdot 0.9336 \), the work done is approximately \( 6,160,320 \mathrm{~J} \).

Key Concepts

Force

Force is a fundamental concept in physics that describes the interaction that can cause an object to change its motion. It is measured in Newtons (N) and is a vector, which means it has both magnitude and direction. Force can act on an object in various ways, such as pushing or pulling.

In the context of this problem, the force is exerted by the towrope pulling the barge. The tension in the rope is given as 12,000 N, which represents the magnitude of the force pulling the barge through the canal. This tension effectively causes the barge to move along the canal's path.

Understanding the direction in which the force is applied is crucial. Since force is a vector, its effectiveness in doing work is influenced by its angle relative to the direction of displacement. This will be elaborated further when discussing the angle of application.

Displacement

Displacement is another fundamental concept in the study of physics, specifically when analyzing work done by a force. It refers to the change in position of an object and is typically measured in meters (m).

In this exercise, the barge is moved through a canal over a distance of 550 meters. This distance represents the displacement of the barge, crucial for calculating the work done. However, it is essential to recognize that displacement is not simply distance; it is a vector quantity like force. This means it also has a direction, from the initial position to the final position.

The significance of displacement lies in its role in calculating work, as it factors into the formula:

• Work Done (W) = Force (F) × Displacement (d) × cos(θ)

Therefore, understanding displacement helps us see how far the force acting on a barge moves it along the intended path.

Angle of application

The angle of application is crucial when analyzing how effective a force is in doing work. This refers to the angle between the force vector and the direction of displacement.

In our scenario, the towrope forms an angle of 21 degrees with the canal bank. Understanding this angle is vital as it affects how much of the tension in the rope contributes effectively to moving the barge. Only the component of force that acts in the direction of displacement does work. The angle modifies this effect through the cosine function in the work formula:

• Work Done (W) = Force (F) × Displacement (d) × \( \cos(\theta) \)

The cosine of the angle, \( \cos(21^{\circ}) \), is approximately 0.9336. This value is a scaling factor for the effective force used in calculating work, bridging the gap between force and displacement directly. Understanding how these angles affect work done is critical in physics, ensuring you consider only the useful component of force.