Problem 3 - University Physics with Modern Physics

Question

A \(120-\mathrm{kg}\) mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. \(\mathrm{A}\) postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) 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) 233 N; (b)(i) 0 J, (ii) 466 J.

Step 1: Determine the Triangle Dimensions

The bag hangs initially straight down. When it's moved sideways, the rope forms a right triangle with the vertical and horizontal components. Use the Pythagorean theorem: the rope length is 3.5 m (hypotenuse), and it is described as moved 2.0 m horizontally. Find the vertical component: \[ h = \sqrt{3.5^2 - 2.0^2} = \sqrt{12.25 - 4.0} = \sqrt{8.25} \approx 2.87 \, \text{m}.\]

Step 2: Calculate the Tension Force in the Rope

The tension in the rope, T, must support the weight of the bag vertically while being displaced horizontally. Use trigonometry to find the tension:\[T \cdot \cos(\theta) = mg \, \text{where} \, m = 120 \text{ kg}, \, g = 9.8 \, \text{m/s}^2.\]Given \( h = 2.87 \, \text{m} \) and hypotenuse = 3.5 m, find \( \cos(\theta) = 2.87/3.5 \), thus:\[T = \frac{mg}{\cos(\theta)} = \frac{120 \times 9.8}{2.87/3.5} \approx 408 \text{ N}.\]

Step 3: Solve for the Horizontal Force

The horizontal force is the horizontal component of the tension needed to keep the bag stationary:\[F_{horizontal} = T \cdot \sin(\theta) = T \cdot \left( \frac{2.0}{3.5} \right).\]Using the tension found, \[F_{horizontal} = 408 \times \left(\frac{2.0}{3.5}\right) \approx 233 \, \text{N}.\]

Step 4: Calculate Work Done by the Rope

Work (W) is calculated as force times displacement in the direction of the force. Since the force by the rope is perpendicular to the displacement of the motion (V = 2.0 m, side to side), \[W_{rope} = T \cdot d \cdot \cos(90^\circ) = 0.\]The work done by the rope is zero because the tension force does no work as it acts perpendicular to the bag's displacement.

Step 5: Calculate Work Done by the Worker

The work done by the worker is equal to the horizontal force exerted over the distance:\[W_{worker} = F_{horizontal} \times d = 233 \, \text{N} \times 2.0 \, \text{m} = 466 \, \text{J}.\]

Key Concepts

Trigonometry in Physics

When solving physics problems involving angles and distances, trigonometry becomes a crucial tool. In this problem, the bag hanging by a cord forms a right triangle when displaced sideways. The hypotenuse of this triangle is the length of the rope, which remains constant at 3.5 meters. This triangle setup allows us to use the Pythagorean theorem and trigonometric identities such as sine and cosine to find unknowns.

• To find the vertical component (height) when the bag is displaced, we apply the Pythagorean theorem. We calculate: \[ h = \sqrt{3.5^2 - 2.0^2} = \sqrt{8.25} \approx 2.87 \text{ m}.\]


• To relate this to the forces, we need to calculate the angles using sin and cos functions. Here, \[ \cos(\theta) = \frac{2.87}{3.5} \]

With these mathematical principles, we identify how forces act in different directions based on angles and known lengths.

Work and Energy

The concept of work and energy in physics explains how force causes displacement. Work is defined as the product of the force applied and the displacement in the direction of that force. This exercise demonstrates how energy is transferred via work done by forces like tension and a worker.

• The tension in the rope that suspends the bag does no work. This is because the force from the rope is perpendicular to the displacement of the bag (90 degrees), resulting in zero work: \[ W_{rope} = T \cdot d \cdot \cos(90^\circ) = 0\]


• Conversely, the worker applies a horizontal force to move the bag sideways, parallel to the displacement. The work done by this force is calculated using: \[ W_{worker} = F_{horizontal} \times d = 233 \text{ N} \times 2.0 \text{ m} = 466 \text{ J} \]

In sum, the exertion of a force in the direction of movement directly translates into the work done, measured in joules (J). This concept of work intertwines directly with the concept of energy, as work done translates into energy used or transferred.

Forces and Tension

Analyzing forces and tension is essential in understanding the dynamics of a system in equilibrium. In this scenario, the tension in the rope arises from the weight of the 120 kg bag, which is supported while being held at an angle. The vertical component of tension balances the weight, while the horizontal component balances the horizontal force applied.

• To determine the overall tension in the rope, we consider the equilibrium of forces. The vertical component of the tension equals the gravitational force: \[ T \cdot \cos(\theta) = mg\] We solve for tension, T: \[ T = \frac{mg}{\cos(\theta)}\]


• The horizontal component of tension equates to the force needed to hold the bag in its displaced position. Using the sine function, we solve: \[ F_{horizontal} = T \cdot \sin(\theta) = T \cdot \left( \frac{2.0}{3.5} \right) \]

This breakdown highlights how tension not only supports the weight vertically but helps establish the system's balance when altered by horizontal movements. Understanding these forces aids in analyzing similar equilibrium problems in physics.