Question: Angle at joint C 0= tan 10 45 10 At joint C, Vertical Fy=0 F+Fsin 45 =0 CD BC 2+FCD (sin 45)=0 F =

Angle at joint C 0= tan 10 45 10 At joint C,

Angle at joint C 0= tan 10 45 10 At joint C, Vertical Fy=0 F+Fsin 45 =0 CD BC 2+FCD (sin 45)=0 F = 2.707 Kips CD At joint C Horizontal F cos45F CD F CE = F = CE - F = CE AC 0 10.003 2.707(cos 45 ) 8.09 Kips

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