= A scissors truss supports three forces at joints B, C, and D as shown in the figure below. The forces have magnitudes F₁ 35 kip, F₂ = 10 kip, and F3 25 kip. The truss is held in equilibrium by a pin at point A and a roller at point E. Note that sides ABC, CDE, AFD, and BFE are straight lines, and member CF is vertical. Joint F lies s = 12 ft to the right and h = 4 ft above joint A, and joint C lies h = 4 ft above joint F. Use the method of joints to determine the force in each member of the truss, and indicate whether they are in tension or compression. Note: Express tension forces as positive (+) and compression forces as negative (-).

Elements Of Electromagnetics
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**Scissors Truss Analysis**

A scissors truss supports three forces at joints B, C, and D as shown in the diagram. The forces have magnitudes \( F_1 = 35 \) kip, \( F_2 = 10 \) kip, and \( F_3 = 25 \) kip. The truss is held in equilibrium by a pin at point A and a roller at point E. It is important to note that sides ABC, CDE, AFD, and BFE are straight lines, and member CF is vertical. Joint F is positioned \( s = 12 \) ft to the right and \( h = 4 \) ft above joint A, while joint C is located \( h = 4 \) ft above joint F. The **method of joints** should be used to determine the forces in each member of the truss, indicating whether they are in tension or compression.

**Note:** Express tension forces as positive (+) and compression forces as negative (-).

**Diagram Explanation**

- The truss is pinned at joint A and supported by a roller at joint E. 
- Forces \( F_1 \), \( F_2 \), and \( F_3 \) act downwards at joints B, C, and D respectively. 
- Joint C lies vertically above joint F. 
- The horizontal distance between joint A and E is \( 2s \) (24 ft), with each \( s \) being 12 ft.
- The height from joint A to joint F and from joint F to joint C is \( h = 4 \) ft each.

This set-up requires analyzing each joint to solve for the internal forces while maintaining the equilibrium of the truss system.
Transcribed Image Text:**Scissors Truss Analysis** A scissors truss supports three forces at joints B, C, and D as shown in the diagram. The forces have magnitudes \( F_1 = 35 \) kip, \( F_2 = 10 \) kip, and \( F_3 = 25 \) kip. The truss is held in equilibrium by a pin at point A and a roller at point E. It is important to note that sides ABC, CDE, AFD, and BFE are straight lines, and member CF is vertical. Joint F is positioned \( s = 12 \) ft to the right and \( h = 4 \) ft above joint A, while joint C is located \( h = 4 \) ft above joint F. The **method of joints** should be used to determine the forces in each member of the truss, indicating whether they are in tension or compression. **Note:** Express tension forces as positive (+) and compression forces as negative (-). **Diagram Explanation** - The truss is pinned at joint A and supported by a roller at joint E. - Forces \( F_1 \), \( F_2 \), and \( F_3 \) act downwards at joints B, C, and D respectively. - Joint C lies vertically above joint F. - The horizontal distance between joint A and E is \( 2s \) (24 ft), with each \( s \) being 12 ft. - The height from joint A to joint F and from joint F to joint C is \( h = 4 \) ft each. This set-up requires analyzing each joint to solve for the internal forces while maintaining the equilibrium of the truss system.
```plaintext
P_AB = number (rtol=0.01, atol=1e-05) kip
P_AF = number (rtol=0.01, atol=1e-05) kip
P_BC = number (rtol=0.01, atol=1e-05) kip
P_BF = number (rtol=0.01, atol=1e-05) kip
P_CD = number (rtol=0.01, atol=1e-05) kip
P_CF = number (rtol=0.01, atol=1e-05) kip
P_DE = number (rtol=0.01, atol=1e-05) kip
P_DF = number (rtol=0.01, atol=1e-05) kip
P_EF = number (rtol=0.01, atol=1e-05) kip
```

Each entry represents a force measurement in kilopounds (kip) with a relative tolerance (`rtol`) of 0.01 and an absolute tolerance (`atol`) of 1e-05. The variables P_AB, P_AF, etc., likely represent forces in different sections or connections within a structural model or system.
Transcribed Image Text:```plaintext P_AB = number (rtol=0.01, atol=1e-05) kip P_AF = number (rtol=0.01, atol=1e-05) kip P_BC = number (rtol=0.01, atol=1e-05) kip P_BF = number (rtol=0.01, atol=1e-05) kip P_CD = number (rtol=0.01, atol=1e-05) kip P_CF = number (rtol=0.01, atol=1e-05) kip P_DE = number (rtol=0.01, atol=1e-05) kip P_DF = number (rtol=0.01, atol=1e-05) kip P_EF = number (rtol=0.01, atol=1e-05) kip ``` Each entry represents a force measurement in kilopounds (kip) with a relative tolerance (`rtol`) of 0.01 and an absolute tolerance (`atol`) of 1e-05. The variables P_AB, P_AF, etc., likely represent forces in different sections or connections within a structural model or system.
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