Before a member can be cut, the reaction loads at the supports need to be determined. Since both members AB and CB are two force members, the direction of the reaction force at A and C are known. Considering the full structure, summation of forces at C gives
ΣMC = 0
-15(10) + FAB cos45 = 0
FAB = 212.1 N
Member Cut at a-a
Member AB can now be cut member at a-a. At the cut, both the shear and moment internal forces are unknowns. The vertical and horizontal components of the force FAB are
FAB-x = FAB-y = FAB / cos45 = 150 N
Vertical and horizontal distances are shown on diagram. Summing moments at the cut gives,
ΣMcut = 0
M + 150(10)(1-cos25) - 150(10)(sin25) = 0
M = 493.4 N
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