∑M_C = 0 (on left part, moments about C): - M_A (clockwise? sign convention: counterclockwise positive) - V_A (up) at distance 3 m from C → moment = - V_A * 3 (clockwise negative) - H_A = 10 kN -> acts at top of column? No, H_A at base A, but horizontal force transmits? Simpler: horizontal forces: H_A (10 kN left) at base, and F (10 kN right) at mid-height. Their moment about C: H_A * (height 4 m) + F * (height 2 m)? No — careful: C is at top of column? No, C is in beam, so height from A to beam = 4 m. Horizontal forces: H_A (10 kN to left) at base, F = 10 kN to right at 2 m high. Moment about C = (H_A * 4 m) clockwise? Let’s do sign: H_A (left) tends to rotate column clockwise around C? Yes: force left at base, center at C above: moment = +H_A 4 (clockwise positive) F (right at 2m high): moment about C = -F * (4-2)= -F 2 (counterclockwise) So net horizontal moment = 10 4 -10 2 = 40-20=20 kNm clockwise (positive). - q resultant 24 kN at 1.5 m from C (left) → moment = -24 1.5= -36 kNm - P=15 kN at 1 m from C (left) → moment = -15 1= -15 kNm - V_A: up at 3 m from C → moment = -V_A*3 - M_A: unknown, assume positive counterclockwise → moment about C = -M_A (because moving A to C, M_A acts counterclockwise at A → at C, it’s clockwise? Let’s keep simple: ∑M_C = 0 → sum = 0: +20 (from horizontals) -36 -15 -3V_A - M_A = 0 → -3V_A - M_A -31 = 0 → 3V_A + M_A = -31 …(3)

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