### Problem: Determine the Tension in the Right CableA wrecking ball is suspended below two cables, each at an angle with the horizontal. The left cable has tension \( T_1 \) and forms an angle of 36 degrees with the horizontal. The right cable has tension \( T_2 \) and forms an angle of 21 degrees with the horizontal. The wrecking ball has a mass of 140 kg.Find the magnitude of tension \( T_2 \).#### Diagram Explanation:The figure provides a visual representation of the problem. Key details in the diagram include:1. **Wrecking Ball and Mass**: - The wrecking ball is depicted as a circular object suspended by two cables. - The mass of the wrecking ball is given as 140 kg.2. **Cables and Tensions**: - There are two cables connected to a horizontal structure, each holding the wrecking ball. - The left cable, with tension \( T_1 \), forms an angle \( \alpha \) of 36 degrees with the horizontal. - The right cable, with tension \( T_2 \), forms an angle \( \beta \) of 21 degrees with the horizontal. 3. **Angles and Forces**: - The left cable (angle \( \alpha \)) and right cable (angle \( \beta \)) have their respective directions and angles labeled in the diagram.#### Solution Steps:To find the tension \( T_2 \), we can use the equilibrium conditions for the wrecking ball, considering both horizontal and vertical components of the forces.1. **Vertical Components**: The sum of the vertical components of the tensions must equal the weight of the wrecking ball. \[ T_1 \sin(36^\circ) + T_2 \sin(21^\circ) = mg \] Where \( m \) is the mass of the wrecking ball (140 kg), and \( g \) is the acceleration due to gravity (approx. 9.81 m/s²).2. **Horizontal Components**: The sum of the horizontal components of the tensions must cancel out. \[ T_1 \cos(36^\circ) = T_2 \cos(21^\circ) \] Solving these equations simultaneously will yield the magnitudes of \( T_1 \) and \( T

We need to find the magnitude of tension T2.

A wracking ball is suspended below two cables each at an angle with the horizontal. The left cable has tension T1 and angle 36 degrees with the horizontal. The right cable has tension T2 and angle 21 degrees with the horizontal. The Wracking ball has mass 140 kg. Find the magnitude of tension T2. T2

A wracking ball is suspended below two cables each at an angle with the horizontal. The left cable has tension T1 and angle 36 degrees with the horizontal. The right cable has tension T2 and angle 21 degrees with the horizontal. The Wracking ball has mass 140 kg. Find the magnitude of tension T2. T2

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Newton’s Third Law of Motion

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### Problem: Determine the Tension in the Right Cable

A wrecking ball is suspended below two cables, each at an angle with the horizontal. The left cable has tension \( T_1 \) and forms an angle of 36 degrees with the horizontal. The right cable has tension \( T_2 \) and forms an angle of 21 degrees with the horizontal. The wrecking ball has a mass of 140 kg.

Find the magnitude of tension \( T_2 \).

#### Diagram Explanation:

The figure provides a visual representation of the problem. Key details in the diagram include:

1. **Wrecking Ball and Mass**:
- The wrecking ball is depicted as a circular object suspended by two cables.
- The mass of the wrecking ball is given as 140 kg.

2. **Cables and Tensions**:
- There are two cables connected to a horizontal structure, each holding the wrecking ball.
- The left cable, with tension \( T_1 \), forms an angle \( \alpha \) of 36 degrees with the horizontal.
- The right cable, with tension \( T_2 \), forms an angle \( \beta \) of 21 degrees with the horizontal.

3. **Angles and Forces**:
- The left cable (angle \( \alpha \)) and right cable (angle \( \beta \)) have their respective directions and angles labeled in the diagram.

#### Solution Steps:

To find the tension \( T_2 \), we can use the equilibrium conditions for the wrecking ball, considering both horizontal and vertical components of the forces.

1. **Vertical Components**:
The sum of the vertical components of the tensions must equal the weight of the wrecking ball.
\[
T_1 \sin(36^\circ) + T_2 \sin(21^\circ) = mg
\]
Where \( m \) is the mass of the wrecking ball (140 kg), and \( g \) is the acceleration due to gravity (approx. 9.81 m/s²).

2. **Horizontal Components**:
The sum of the horizontal components of the tensions must cancel out.
\[
T_1 \cos(36^\circ) = T_2 \cos(21^\circ)
\]

Solving these equations simultaneously will yield the magnitudes of \( T_1 \) and \( T

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