Problem 28 - Essential Calculus Early Transcendentals

Question

Ropes 3 \(\mathrm{m}\) and 5 \(\mathrm{m}\) in length are fastened to a holiday deco- ration that is suspended over a town square. The decoration has a mass of 5 \(\mathrm{kg}\) . The ropes, fastened at different heights, make angles of \(52^{\circ}\) and \(40^{\circ}\) with the horizontal. Find the tension in each wire and the magnitude of each tension.

Step-by-Step Solution

Answer

The tensions are approximately 37.88 N in the 3 m rope and 30.61 N in the 5 m rope.

Step 1: Understand the Problem

We need to find the tensions in two ropes holding a decoration. The decoration weighs 5 kg and is held at angles of 52° and 40° with the horizontal by 3 m and 5 m ropes. Since the decoration is not moving, the forces on it must balance.

Step 2: Force Analysis

Draw a free-body diagram to show the forces. The weight of the decoration acts downward with a force \(F_w = mg = 5 \times 9.8 = 49 \text{ N}\). Let \(T_1\) and \(T_2\) be the tensions in the 3 m and 5 m ropes, respectively. Calculate the vertical and horizontal components of each tension using trigonometry. The vertical components sum to the weight, and the horizontal components balance each other.

Step 3: Set up Equations

Use the angles and known weight to set up the system of equations: \(T_1 \cdot \sin(52°) + T_2 \cdot \sin(40°) = 49\) for vertical components, and \(T_1 \cdot \cos(52°) = T_2 \cdot \cos(40°)\) for horizontal balance.

Step 4: Solve for One Tension in Terms of the Other

From the horizontal component equation: \(T_1 \cdot \cos(52°) = T_2 \cdot \cos(40°)\), solve for \(T_1\) in terms of \(T_2\): \(T_1 = T_2 \cdot \frac{\cos(40°)}{\cos(52°)}\).

Step 5: Substitute and Solve for Tensions

Substitute \(T_1\) from Step 4 into the vertical equation: \(T_2 \cdot \frac{\cos(40°)}{\cos(52°)} \cdot \sin(52°) + T_2 \cdot \sin(40°) = 49\). Solve this equation to find \(T_2\), then use \(T_1 = T_2 \cdot \frac{\cos(40°)}{\cos(52°)}\) to find \(T_1\).

Step 6: Calculate the Actual Tensions

Evaluate the trigonometric functions and solve the equations: \(T_1 ≈ 37.88 \text{ N}\) and \(T_2 ≈ 30.61 \text{ N}\). Ensure your calculator is in degree mode for these calculations.

Key Concepts

Free-Body Diagram

When analyzing forces in physics, a free-body diagram is a useful tool. It visually represents all the forces acting on an object, allowing us to understand the problem better. Imagine a diagram where a decoration is hanging from two ropes.
Each rope exerts a force or tension on the decoration, represented by arrows in the diagram. These forces act at angles to the horizontal, aligned with the ropes at 52° and 40° respectively.
The forces in a free-body diagram help us isolate and solve for unknown variables, like the tension in each rope. The weight of the decoration, acting downward, is another crucial force in this scenario, calculated by multiplying the mass by gravity, which is 9.8 m/s², resulting in 49 N. This weight needs to be balanced by the vertical components of the tensions in the ropes to keep the decoration stationary.

Trigonometry in Physics

Trigonometry often comes to the rescue when solving problems with angles in physics. The ropes make specific angles with the horizontal, and we use trigonometry to break down these tension forces into components.
For tension forces, we usually resolve them into horizontal and vertical components using the sine and cosine functions. The vertical component helps balance the weight of the object suspended by the ropes. The equations we come up with using trigonometry are:

• Vertical: \( T_1 \cdot \sin(52°) + T_2 \cdot \sin(40°) = 49 \)
• Horizontal: \( T_1 \cdot \cos(52°) = T_2 \cdot \cos(40°) \)

With these equations, we can solve for unknown tensions using trigonometric identities.
Trigonometry bridges the gap between geometric perspectives and algebra, helping us make sense of forces at angles.

Forces in Equilibrium

When an object is stationary, like our holiday decoration, it is at equilibrium. This means the net force acting on it is zero. For equilibrium in our context, we have both vertical and horizontal forces to consider.
Vertically, the sum of the upward components of tensions in the ropes must equal the downward force due to the weight of the decoration. Horizontally, the components of the tensions must balance each other out to ensure there is no movement sideways.
The concept of equilibrium assures that these force components can be logically set in equations:

• Vertically: \( T_1 \cdot \sin(52°) + T_2 \cdot \sin(40°) = 49 \)
• Horizontally: \( T_1 \cdot \cos(52°) = T_2 \cdot \cos(40°) \)

By solving these equations, we determine the needed tensions (\(T_1 \approx 37.88 \text{ N}\) and \(T_2 \approx 30.61 \text{ N}\)) to maintain equilibrium. Understanding equilibrium is essential in physics to solve many real-world problems where balance is crucial.