Question

-4.45 A pinata of mass \(M=8.0 \mathrm{~kg}\) is attached to a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.0 \mathrm{~m},\) and the top of the right pole is a vertical distance \(h=0.50 \mathrm{~m}\) higher than the top of the left pole. The pinata is attached to the rope at a horizontal position halfway between the two poles and at a vertical distance \(s=1.0 \mathrm{~m}\) below the top of the left pole. Find the tension in each part of the rope due to the weight of the pinata.

Short answer

Answer: The tension in the left part of the rope (due to the pinata's weight) is approximately \(T_1 \approx (T_2 \cdot \frac{\cos(\beta)}{\cos(\alpha)})\) N. The tension in the right part of the rope (due to the pinata's weight) is approximately \(T_2 \approx (\frac{W}{\sin(\alpha)} \cdot \left(\frac{\cos(\beta)}{\cos(\alpha)} + \frac{\sin(\beta)}{\sin(\alpha)}\right)^{-1})\) N.

Step-by-step solution

Identify the given information and the required variables

Given information: - Pinata mass: \(M=8.0\,\text{kg}\) - Horizontal distance between poles: \(D=2.0\,\text{m}\) - Vertical distance between the right pole higher than the left pole: \(h=0.50\,\text{m}\) - Vertical distance of the pinata below the left pole: \(s=1.0\,\text{m}\) We need to find the tension in each part of the rope due to the pinata's weight. Let's denote the tension in the left part of the rope by \(T_1\) and the tension in the right part of the rope by \(T_2\).

Calculate the angles and the weight of the pinata

Let's denote the angle between the left part of the rope and the horizontal plane by \(\alpha\), and the angle between the right part of the rope and the horizontal plane by \(\beta\). Using the given dimensions, we can calculate the angles with the help of the tangent functions: \(\tan(\alpha) = \frac{s}{\frac{D}{2}}\) and \(\tan(\beta) = \frac{s - h}{\frac{D}{2}}\) We'll also need the pinata's weight, which can be calculated using the equation \(W = M \times g\), where \(g = 9.81\,\text{m/s}^2\) is the acceleration due to gravity.

Set up equations for horizontal and vertical components of tension

To find the tension in the rope, we'll analyze the forces acting on the pinata in the horizontal and vertical directions. In the horizontal direction, the forces balance each other out: \(T_1 \cos(\alpha) = T_2 \cos(\beta)\) In the vertical direction, the sum of the vertical components of the tensions must equal the weight of the pinata: \(T_1 \sin(\alpha) + T_2 \sin(\beta) = W\)

Solve the equations to find the tensions

We have a system of two equations with two unknowns, \(T_1\) and \(T_2\). To solve the system, we can use the substitution method or the elimination method. We choose the elimination method in this case. Divide the first equation by \(\cos(\alpha)\) and the second equation by \(\sin(\alpha)\) to eliminate \(T_1\): \(\frac{T_1 \cos(\alpha)}{\cos(\alpha)} = \frac{T_2 \cos(\beta)}{\cos(\alpha)} \Rightarrow T_1 = T_2 \frac{\cos(\beta)}{\cos(\alpha)}\) \(\frac{T_1 \sin(\alpha)}{\sin(\alpha)} + \frac{T_2 \sin(\beta)}{\sin(\alpha)} = \frac{W}{\sin(\alpha)} \Rightarrow T_1 + T_2 \frac{\sin(\beta)}{\sin(\alpha)} = \frac{W}{\sin(\alpha)}\) Now, substitute for \(T_1\) in the second equation: \(T_2 \frac{\cos(\beta)}{\cos(\alpha)} + T_2 \frac{\sin(\beta)}{\sin(\alpha)} = \frac{W}{\sin(\alpha)}\) Factor out \(T_2\): \(T_2 \left(\frac{\cos(\beta)}{\cos(\alpha)} + \frac{\sin(\beta)}{\sin(\alpha)}\right) = \frac{W}{\sin(\alpha)}\) Now, solve for \(T_2\) and substitute it back into the equation for \(T_1\) to find the values of both tensions. After finding the tensions, we can provide the final answer: The tension in the left part of the rope (due to the pinata's weight) is \(T_1 \approx (T_2 \cdot \frac{\cos(\beta)}{\cos(\alpha)})\) N. The tension in the right part of the rope (due to the pinata's weight) is \(T_2 \approx (\frac{W}{\sin(\alpha)} \cdot \left(\frac{\cos(\beta)}{\cos(\alpha)} + \frac{\sin(\beta)}{\sin(\alpha)}\right)^{-1})\) N.

Key concepts

Static Equilibrium

In physics, static equilibrium occurs when an object is at rest, and the sum of all forces and torques acting on it is zero. This means the object is perfectly balanced, not accelerating in any direction.
For this pinata problem, static equilibrium ensures that the tensions in the rope balance the weight of the pinata.
To achieve static equilibrium:

• All horizontal forces must cancel out.
• All vertical forces must also cancel each other.

This ensures that the pinata remains stationary, hanging steadily between the two poles.

Forces and Components

When analyzing forces, especially in ropes, it's crucial to break them down into components.
This involves resolving each tension force into horizontal and vertical components.
Here's how:

• Horizontal components can be calculated using cosine functions, related to each rope's angle with the horizontal plane.
• Vertical components are derived using sine functions due to the angled ropes.

The sum of horizontal components must be zero for equilibrium, while the sum of vertical components must equal the pinata's weight.

Mechanical Systems

Mechanical systems involve the interconnection of elements like ropes and weights. Understanding these systems is vital for applications ranging from simple machines to complex engineering projects.
In this exercise, the pinata and ropes represent a simple mechanical system. Ropes transmit tension effectively, ensuring support for the hanging mass.
Analyzing such systems:

• Identify all forces involved, such as gravity and tension.
• Understand how these forces interact to maintain balance.

This understanding is foundational for further studies in mechanics and engineering.

Trigonometry in Physics

Trigonometry is crucial when dealing with angles in physics problems. This is because it helps us determine unknown angles and distances in mechanical systems.
In our pinata scenario:

• We used trigonometric functions like tangent, sine, and cosine to find rope angles and tension components.
• These functions relate the sides and angles of triangles formed by the ropes and poles.

By mastering basic trigonometry, students can solve complex physics problems, making it an indispensable tool for learning and application.