Consider an arbitrary truss structure composed of diagonal, vertical, and horizontal members fixed to the wall. To calculate the force acting on members CB, GB, and GH, method of sections can be used. The loads and lengths of the horizontal and vertical members are known parameters, as shown in the figure.

To begin, a cut is made along a plane intersecting CB, GB, and GH members, and a free-body diagram of the right side section is drawn.

The moment equilibrium equation about point G is applied.

The result gives the force along member CB as 4 kN, with a positive sign indicating tension in the member.

Now, F_CB can be expressed using a slope triangle in BCG, while FGH can be expressed using a slope triangle in CEG. Considering the summation of vertical and horizontal forces, the force equilibrium equations can be written as the following:

The value of F_CB is substituted, and the force equilibrium equations are solved simultaneously. The result yields the force F_GH as -7.454 kN and F_GB as 3.772 kN. The negative sign of F_GH indicates that the force is compressive, while the positive sign for F_GB indicates the tensile force.