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P2829key(2)

# P2829key(2) - P 28 FBD 2 pts 8 ft = 2.438 m Ma = 0 14kN...

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P 28 FBD: 2 pts 8 ft 2.438m = Σ Ma = 0 14kN 2.438 m 36kN 2.438 m + Cy 3 2.438 m - 0 = Cy 14kN 2.438 m 36kN 2.438 m + 3 2.438 m = Cy 16.67 kN = 3 pts Σ Fy = 0 Cy 14kN - A sin 135deg ( ) + 0 = 16.67kN 14kN - A sin 135deg ( ) + 0 = A 2.67 kN sin 135.0 deg ( ) - = A 3.78 - kN = 3 pts Σ Fx = 0 Cx 36 kN - A cos 135deg ( ) + 0 = Cx 36 kN - 3.78 - kN ( ) cos 135deg ( ) + 0 = Cx 36.0 kN 3.78 kN cos 135 deg ( ) + = Cx 33.33 kN = 3 pts Alternate version: FBD 2 pts Σ Fx = 0 Cx 36 kN - 0 = Cx 36kN = 3 pts Σ Ma = 0 14kN 2.438 m 36kN 2.438 m + Cy 3 2.438 m - 0 = Cy 14kN 2.438 m 36kN 2.438 m + 3 2.438 m = Cy 16.67 kN = 3 pts Σ Fy = 0 Cy 14kN - Ay + 0 = 16.67kN 14kN - Ay + 0 = Ay 14kN 16.67kN - = Ay 2.67 - kN = 3 pts

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P 29 Assume that the unknowns are in tension! cos 22deg ( ) 0.927 = sin 35deg ( ) 0.574 = cos 35deg ( ) 0.819 = sin 22deg ( ) 0.375 = FBD: 2 pts Σ Fx = 0 Fab - cos 35deg ( ) Fbc cos 22deg ( ) - 0 = Fab - 0.819 Fbc 0.927 - 0 = Fab 1.132 - Fbc = 1lbf 4.448N = Σ Fy = 0 500lbf 2224 N = 500 - lbf Fab sin 35deg ( ) - Fbc sin 22deg ( ) - 0 = 500 - lbf 1.132 - Fbc ( ) 0.574 - Fbc 0.375 - 0 = 500 - lbf 0.650 Fbc + 0.375 Fbc
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P2829key(2) - P 28 FBD 2 pts 8 ft = 2.438 m Ma = 0 14kN...

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