Question

Solve by applying the stationary principle of potential energy.
Find internal forces and reactions

(a) Figura 1-6 6m 15k 15k 4m Am 4m

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Answer #1

B15 15t15- 45 k (1) BrtRA=0 EMB6 RAX615x4 4-15x-15x12 20. RA- 60k )) at joint Stmo,-2 FP S . И 15k Cose X_Q 15 0 ension) f 33at joint E PED FEC 15k FED Sint5-15 = 0 21-21 k(T) oTEDroses- FEC0 Or 30-21-2 x-e FEC45kC) at joint FEA EE FeA=4SK) fAB FAD a

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