In the figure below, calculate the forces F1 , F2 , F3 , H2 , V2 ,V3 using the inverse of the coefficient matrix. 1000 lb 90 F3 60 30 H 2 3. t Tip: For the static balance of the frame, you can use the equations between the forces given on points 1, 2 and 3 in the horizontal and vertical directions, respectively. EFH = 0 = F, cos 30°- Fa cos 60° EFv = 0= F sin 30° + F3 sin 60° - 1000 EFu = 0 = -F,- F, cos 30° + H2 EFv = 0 = -F sin 30° + V2 %3D EFH = 0 = F2+ F3 cos 60° EFv = 0 = -F3 sin 60° + V3 %3D
In the figure below, calculate the forces F1 , F2 , F3 , H2 , V2 ,V3 using the inverse of the coefficient matrix. 1000 lb 90 F3 60 30 H 2 3. t Tip: For the static balance of the frame, you can use the equations between the forces given on points 1, 2 and 3 in the horizontal and vertical directions, respectively. EFH = 0 = F, cos 30°- Fa cos 60° EFv = 0= F sin 30° + F3 sin 60° - 1000 EFu = 0 = -F,- F, cos 30° + H2 EFv = 0 = -F sin 30° + V2 %3D EFH = 0 = F2+ F3 cos 60° EFv = 0 = -F3 sin 60° + V3 %3D
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T5)
Transcribed Image Text:In the figure below, calculate the forces
F1 , F2 , F3 , H2 , V2 , V3 using the inverse of the coefficient
matrix.
1000 lb
90
60
30
H 2
3
Tip: For the static balance of the frame, you can use the equations between the forces
given on points 1, 2 and 3 in the horizontal and vertical directions, respectively.
EFH = 0= F, cos 30°- Fa cos 60°
EFv = 0= F, sin 30° + F3 sin 60°- 1000
%3D
EFu = 0 = -F, - F cos 30°+ H2
EF = 0 = -F, sin 30° + V2
%3D
£FH =0 = F2+ F cos 60°
EFv = 0 =-F3 sin 60° + V3
%3D
%3D
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