Consider a hydraulic hoist supporting a load of 1 kN. Assuming a simplified schematic representation of this frame structure, the force acting on BD and BF members can be determined.

The member dimensions and the weight of the load are known parameters. This structure can be considered a frame with BF and BD acting as two-force members, while EFG and EDC are multi-force members.

A free-body diagram for the member EFG is considered. The inclined force F_BF can be resolved into the vertical and horizontal components. The moment equilibrium condition is applied to joint E.

The force F_BF is calculated to be 1.546 kN. Then, the horizontal force equilibrium condition is applied at joint E.

Substituting the value of the horizontal component of F_BF into the equilibrium equation, the horizontal reaction force at E is calculated as 0.375 kN. Similarly, the vertical force equilibrium condition is applied.

The calculated vertical reaction force at joint E is 0.500 kN.

Now, a free-body diagram for the member EDC is considered. The horizontal and vertical components of FBD can be expressed using a slope triangle. The moment equilibrium condition at point C is applied.

The force F_BD is calculated as 1.677 kN.