Can you please solve number two The image shows a mechanical problem involving a boom, cables, and a load. The boom OA carries a load $ P $ and is supported by two cables. The system is defined as follows:1. The tension in cable AB is 183 lb.2. The resultant of the load $ P $ and the forces exerted at point A by the two cables must be directed along OA.The problem consists of two parts:a) Determine the tension in cable AC.b) Determine the magnitude of the load $ P $.**Diagrams Explanation:**There are two diagrams provided:1. Top View: - The boom is shown extending along the x-axis. - Cable AB is attached to the boom at an angle of 40° to the horizontal plane. - Cable AC is similarly attached but is situated in the z-direction, making angles with the x-axis and z-axis.2. Side View: - This diagram provides the lengths and angles of the cables. - Cable AB connects to point B, located 24 in. along the z-axis and 29 in. along the x-axis from point A. - Cable AC connects to point C, situated 36 in. along the y-axis and 25 in. along the z-axis from point A. - Point A is connected to the load $ P $, positioned vertically downward, 48 in. beneath the z-plane.The objective is to calculate the unknown tensions and the load magnitude by resolving the forces within the system.

The diagram of force vectors is given as, The co-ordinate at point A,The co-ordinate at point B,The…

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Can you please solve number two

The image shows a mechanical problem involving a boom, cables, and a load. The boom OA carries a load $ P $ and is supported by two cables. The system is defined as follows:

1. The tension in cable AB is 183 lb.
2. The resultant of the load $ P $ and the forces exerted at point A by the two cables must be directed along OA.

The problem consists of two parts:
a) Determine the tension in cable AC.
b) Determine the magnitude of the load $ P $.

**Diagrams Explanation:**

There are two diagrams provided:

1. Top View:
- The boom is shown extending along the x-axis.
- Cable AB is attached to the boom at an angle of 40° to the horizontal plane.
- Cable AC is similarly attached but is situated in the z-direction, making angles with the x-axis and z-axis.

2. Side View:
- This diagram provides the lengths and angles of the cables.
- Cable AB connects to point B, located 24 in. along the z-axis and 29 in. along the x-axis from point A.
- Cable AC connects to point C, situated 36 in. along the y-axis and 25 in. along the z-axis from point A.
- Point A is connected to the load $ P $, positioned vertically downward, 48 in. beneath the z-plane.

The objective is to calculate the unknown tensions and the load magnitude by resolving the forces within the system.

Expert Solution

Step 1: Write the coordinates and postion vector-

The diagram of force vectors is given as,

The co-ordinate at point A,

$A equals open parentheses 48 comma 0 comma 0 close parentheses$

The co-ordinate at point B,

$B equals open parentheses 0 comma 29 comma 24 close parentheses$

The co-ordinate at point C,

$C equals open parentheses 0 comma 25 comma negative 36 close parentheses$

The vector AB,

$table row cell A B end cell equals cell open parentheses 0 minus 48 close parentheses i plus open parentheses 29 minus 0 close parentheses j plus open parentheses 24 minus 0 close parentheses k end cell row blank equals cell negative 48 i plus 29 j plus 24 k end cell end table$

The unit vector in the direction of AB,

$lambda subscript A B end subscript equals fraction numerator A B over denominator open vertical bar A B close vertical bar end fraction equals fraction numerator negative 48 i plus 29 j plus 24 k over denominator square root of open parentheses negative 48 close parentheses squared plus open parentheses 29 close parentheses squared plus open parentheses 24 close parentheses squared end root end fraction equals 1 over 61 open parentheses negative 48 i plus 29 j plus 24 k close parentheses$

The force in the direction of AB is expressed as follows-

$table row cell bold italic F subscript bold A bold B end subscript end cell equals cell F subscript A B end subscript lambda subscript A B end subscript end cell row blank equals cell F subscript A B end subscript over 61 open parentheses negative 48 i plus 29 j plus 24 k close parentheses end cell end table$

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