Determine the tensions in cables AB, AC, and AD. B 2.3 1.45 m CO D 2.20 m 1.30 m 0.40 m
Determine the tensions in cables AB, AC, and AD. B 2.3 1.45 m CO D 2.20 m 1.30 m 0.40 m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
![**Determine the tensions in cables \( AB \), \( AC \), and \( AD \).**
The diagram shows a system where three cables \( AB \), \( AC \), and \( AD \) are supporting a mass of \( 180 \, \text{kg} \). The point \( A \) is positioned below the plane containing points \( B \), \( C \), and \( D \). The distances between these points are labeled as follows:
- The vertical distance from \( A \) to the horizontal plane is \( 2.3 \, \text{m} \).
- Horizontal distances:
- From \( B \) to \( C \): \( 1.45 \, \text{m} \)
- From \( C \) to \( D \): \( 2.20 \, \text{m} \)
- From \( B \) to \( D \): \( 1.30 \, \text{m} \)
- From \( C \) to the vertical line down from \( A \): \( 0.40 \, \text{m} \)
**Answers:**
- \( T_{AB} = \) [Input Box] \( \text{N} \)
- \( T_{AC} = \) [Input Box] \( \text{N} \)
- \( T_{AD} = \) [Input Box] \( \text{N} \)
Students are required to calculate the tensions in the cables using static equilibrium equations for the system. The tension in each cable corresponds to the force needed to keep the mass suspended without movement.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb9943cd1-ce44-4500-ac20-6c5cd37adcf9%2Fee38e256-09ec-4e4a-ae5f-0afb1e370024%2Fht8iegj_processed.png&w=3840&q=75)
Transcribed Image Text:**Determine the tensions in cables \( AB \), \( AC \), and \( AD \).**
The diagram shows a system where three cables \( AB \), \( AC \), and \( AD \) are supporting a mass of \( 180 \, \text{kg} \). The point \( A \) is positioned below the plane containing points \( B \), \( C \), and \( D \). The distances between these points are labeled as follows:
- The vertical distance from \( A \) to the horizontal plane is \( 2.3 \, \text{m} \).
- Horizontal distances:
- From \( B \) to \( C \): \( 1.45 \, \text{m} \)
- From \( C \) to \( D \): \( 2.20 \, \text{m} \)
- From \( B \) to \( D \): \( 1.30 \, \text{m} \)
- From \( C \) to the vertical line down from \( A \): \( 0.40 \, \text{m} \)
**Answers:**
- \( T_{AB} = \) [Input Box] \( \text{N} \)
- \( T_{AC} = \) [Input Box] \( \text{N} \)
- \( T_{AD} = \) [Input Box] \( \text{N} \)
Students are required to calculate the tensions in the cables using static equilibrium equations for the system. The tension in each cable corresponds to the force needed to keep the mass suspended without movement.
Expert Solution
Step 1
Force in terms of the cartesian system is obtained by multiplying the magnitude of force and unit vector.
F= F
Hence, force in the cartesian system is F, the magnitude of the force is F, and the unit vector of force is .
Formula to calculate unit vector is :-
Here, the position vector is r, and the magnitude of the position vector is
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