Problem 3 - University Physics with Modern Physics

Question

A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

Step-by-Step Solution

Answer

(a) Horizontal force is needed to hold the bag, calculated using tension components. (b) No work is done by the rope; work done by the worker is calculated over the displacement distance.

Step 1: Analyzing Forces in the System

To find the horizontal force needed to hold the bag in the displaced position, we first need to consider the forces acting on the bag. The weight of the bag acts vertically downward, and the tension in the rope has both a vertical and horizontal component when the bag is displaced. We need to find the horizontal component of the tension.

Step 2: Calculating the Angle of Displacement

The bag swings to form a right triangle where the rope is the hypotenuse. The vertical part of the triangle is given by the rope remaining vertical minus the displacement, calculated as \(3.5 - 2.0 = 1.5 \, \text{m}\). We can find the angle \(\theta\) that the rope makes with the vertical using trigonometry: \(\cos \theta = \frac{1.5}{3.5}\).

Step 3: Determining the Horizontal Tension Component

The tension in the rope, \(T\), balances the weight of the bag: \(T \cdot \cos \theta = mg\). Thus, \(T = \frac{mg}{\cos \theta}\). The horizontal component needed is \(T \cdot \sin \theta\). Substituting the value of \(T\) gives \(F_{\text{horizontal}} = mg \tan \theta\).

Step 4: Calculate the Horizontal Force

Substitute the known values: the mass \(m = 90.0 \, \text{kg}\), and the gravitational acceleration \(g = 9.81 \, \text{m/s}^2\). Hence, \(F_{\text{horizontal}} = 90.0 \times 9.81 \times \tan(\cos^{-1}(\frac{1.5}{3.5}))\). Calculate to find the exact force required.

Step 5: Evaluating Work Done by the Rope

The work done by the tension in the rope is zero because tension acts perpendicular to the direction of motion (i.e., the displacement of the bag), and work is only done when there is a component of force in the direction of displacement.

Step 6: Calculating Work Done by the Worker

The work done by the worker equals the force exerted by the worker multiplied by the distance through which the force acts: \(W = F_{\text{horizontal}} \times 2.0\). Substitute the horizontal force obtained earlier to find the work done by the worker.

Key Concepts

Forces in Physics

In physics, forces are essential for understanding how objects interact with each other. A force can push or pull an object and is usually measured in newtons (N). In our specific exercise, several forces come into play: gravity, tension, and external forces applied by a worker.

Gravity is the force that pulls objects towards the Earth's center. For our mailbag, this force acts downward and can be calculated as the product of the mass of the mailbag and the acceleration due to gravity: \( F_{ ext{gravity}} = mg \), where \( m \) is the mass (90kg in this scenario) and \( g \) is the gravitational acceleration (approximately \( 9.81 \, \text{m/s}^2\)).

The tension in the rope must counteract the weight of the mailbag. When the bag is displaced laterally, this tension has both vertical and horizontal components that work together to maintain balance. Understanding these forces helps in calculating the necessary horizontal force or work done by an external agent like a postal worker.

Tension in Ropes

Tension is a force that is transmitted through a string, cable, or rope when it is pulled tight by forces acting from both ends. In the physics of ropes, tension is considered the same throughout, assuming a massless rope.

In the given problem, the tension in the rope must balance the vertical force, which is the weight of the mailbag. When the bag is moved sideways, the tension adopts an angle to keep the bag in equilibrium. The tension's vertical component acts to balance gravity, while the horizontal component provides resistance to movement.

To find the horizontal component of the tension, we first calculate the tension as \( T = \frac{mg}{\cos \theta} \). Then, to find the horizontal force, we use \( T \sin \theta \). This is crucial for solving problems involving objects suspended by ropes and displaced from their natural position.

Work and Energy

Work and energy are fundamental concepts in physics, interlinked through the idea of forces causing movement. Work is done when a force moves an object over a distance. The formula for work is \( W = Fd \cos \phi \), where \( F \) is the force, \( d \) is the distance moved, and \( \phi \) is the angle between the force and the direction of motion.

In our exercise, calculating work involves two specific cases: work done by the rope and work done by the worker. The rope’s tension always acts perpendicular to the bag’s displacement, meaning no work is done by the rope. By contrast, the worker's effort to hold the bag sideways across a displacement of 2.0 m requires calculating the work done using the horizontal force component: \( W = F_{ ext{horizontal}} \times 2.0 \).

This relates to the principle that energy is transferred when work is done, evident as the worker applies energy to hold the bag in a new position.

Trigonometry in Physics

Trigonometry is an invaluable tool in physics, especially when dealing with components of forces or vectors. It involves the study of relationships in triangles, primarily focusing on the sine, cosine, and tangent functions.

In this exercise, trigonometry is used to dissect the forces acting on the displaced mailbag. The situation forms a right triangle where the rope length is the hypotenuse, and the vertical displacement provides one leg of the triangle.

By using trigonometric functions, particularly the cosine, you can find the angle made with the vertical: \( \cos \theta = \frac{1.5}{3.5} \). Once \( \theta \) is known, you can calculate \( \tan \theta \), helpful in finding the horizontal component of the force. Trigonometry simplifies complex relationships into manageable equations, allowing for precise calculations in physics problems.