Resolve the force F = 900 N acting at B as shown in Fig. into parallel components at O and A.[Figure: horizontal bar OAB; O at the left end, A at 1 m from O, B at the right end, 2 m from A (OB = 3 m). A vertical downward force of 900 N acts at B.]

Given: 900 N (\(\downarrow\)) at B; OA = 1 m, AB = 2 m, OB = 3 m. Let the parallel (vertical) components be \(F_O\) at O and \(F_A\) at A, both assumed downward.

Condition 1 (same resultant): \(F_O + F_A = 900\)

Condition 2 (same moment about O) (Varignon): \(F_A \times 1 = 900 \times 3 \Rightarrow F_A = 2700\ \text{N}\)

Then \(F_O = 900 - 2700 = -1800\ \text{N}\), i.e. 1800 N acting upward.

Check (moment about A): given force: \(900 \times 2 = 1800\) N·m clockwise; components: 1800 N upward at 1 m left of A gives \(1800 \times 1 = 1800\) N·m clockwise. ✔

Answer: \(F_A = 2700\ \text{N}\ (\downarrow)\) at A and \(F_O = 1800\ \text{N}\ (\uparrow)\) at O.