Create 1 Program in Python, to Solve two different Engineering Problems. Your two programs must include the following elements:

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Create 1 Program in Python, to Solve two different Engineering Problems.
Your two programs must include the following elements:
1. Welcome Message
2. Problem Description Message
(background, required inputs & expected results)
3. Set/Capture Input Data
4. Perform the Main Process
(corresponds to the Engineering Problem)
5. Display Results, feedback, and Comments
(Based on results)
6. Goodbye Message
Rules:
- You can select your Engineering Problems from the Engineering Problems Source document
- You must select two different Problems from the source document.
- You can select only available problems that were not already selected by another group,
selection is on first-come, first-serve basis.
- Problems Selection (booking) will be done in the Telegram Group.
- After 7 days, groups that did not select problems, will be assigned problems at random
Bonuses:
- Include Validation (whenever User input is involved)
- Solve Multiple Engineering Problems in one program (user can select from main menu)
- Add export data option to write results/findings into an external document
- Include Plotting/Visualization in your program (using graphics libraries)
- Add main menu options, i.e. select program, export data, plot data, restart, quit, etc.
- Use advanced/new concepts not taught in class
- Any other feature (code readability, good UI, Processing effects, etc.) 

*3-48. Determine the tension developed in cables AB, AC,
and AD required for equilibrium of the 150-kg crate.
B
0.3 m,
0.6 m
0.3 m
40.6 m
0,6 m
0.6 m
[A
0.9 m
D
Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as
(-0.6 – 0)i + (0.3 – 0)j + (0.6 – 0)k
FA = FAB
Fi+ FAnj
FAk
LV-0.6 – 0)² + (0.3 – 0)² + (0.6 – 0)²
(-0.6 – 0)i + (-0.6 – 0)j + (0.3 – 0)k
FAC = FAc
(-0.6 – 0)² + (-0.6 – 0)² + (0.3 – 0)²
FAD = FADİ
W = [-300k] N
Equations of Equilibrium: Equilibrium requires
EF = 0; FAn + FAc + FAp + W = 0
_2,
Fi-Fcj+ Fack + F,i + (-150(9.81)k) = 0
F
-Fk
+
AB
+ FAD
-F
F
AC
j+
F
- 300 k = 0
AB
AC
AB
AB
AC
Equating the i, j, and k components yields
· Fxc+ FAp = 0
(1)
AB
3
· Fxc = 0
(2)
AB
3
AC
FAB +
Fxc – 150(9.81) = 0
AC
3
(3)
Transcribed Image Text:*3-48. Determine the tension developed in cables AB, AC, and AD required for equilibrium of the 150-kg crate. B 0.3 m, 0.6 m 0.3 m 40.6 m 0,6 m 0.6 m [A 0.9 m D Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as (-0.6 – 0)i + (0.3 – 0)j + (0.6 – 0)k FA = FAB Fi+ FAnj FAk LV-0.6 – 0)² + (0.3 – 0)² + (0.6 – 0)² (-0.6 – 0)i + (-0.6 – 0)j + (0.3 – 0)k FAC = FAc (-0.6 – 0)² + (-0.6 – 0)² + (0.3 – 0)² FAD = FADİ W = [-300k] N Equations of Equilibrium: Equilibrium requires EF = 0; FAn + FAc + FAp + W = 0 _2, Fi-Fcj+ Fack + F,i + (-150(9.81)k) = 0 F -Fk + AB + FAD -F F AC j+ F - 300 k = 0 AB AC AB AB AC Equating the i, j, and k components yields · Fxc+ FAp = 0 (1) AB 3 · Fxc = 0 (2) AB 3 AC FAB + Fxc – 150(9.81) = 0 AC 3 (3)
Equating the i, j, and k components yields
2
FAC + FAD = 0
(1)
AB
3
2
F
(2)
AB
3
AC
FAB +
FAC – 150(9.81) = 0
(3)
Solving Eqs. (1) through (3) yields
FA = 1765.8 N
FAc = 882.9 N
FAD = 1765.8 N
Ans
Ans
%3!
Ans
FAB
0.6 m
- 0.3 m
0.3 m
0.6 m
0,6 m
Fre
0.6 m
A
FAD
W = 150(9.81) N
Transcribed Image Text:Equating the i, j, and k components yields 2 FAC + FAD = 0 (1) AB 3 2 F (2) AB 3 AC FAB + FAC – 150(9.81) = 0 (3) Solving Eqs. (1) through (3) yields FA = 1765.8 N FAc = 882.9 N FAD = 1765.8 N Ans Ans %3! Ans FAB 0.6 m - 0.3 m 0.3 m 0.6 m 0,6 m Fre 0.6 m A FAD W = 150(9.81) N
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