Question 3.5.1: A space truss ABCDEF is hinged to a vertical wall in the zx ...

At joint F, all bars except bar 1 lie in the horizontal plane. Hence, by Rule 1, the force in bar 1 is zero, i.e.

$S_{1}=\underline{0}$

At joint E, $S_{6}$, the force in bar 6 is immediately determined from the condition $\sum{P_{z}} = 0:$

$S_{6 }  \cos 60° = P$

Then

$S_{6} =\underline{2P}$      (compressive)

The horizontal component of $S_{6}$ is

$H_{6} = S_{6} \sin 60° = 1.73P$

From Fig. 3.5-1(b), it is clear that

$S_{4} = H_{6}  \cos 60° = \underline{0.87P}$    (tensile)

$S_{5} = H_{6} \sin 60° = \underline{1.5P}$       (tensile)

By symmetry (Fig. 3.5-1(b)),

$S_{3} = H_{6} = \underline{1.73P}$      (compressive)

$S_{2} = S_{4} = \underline{0.87P}$    (tensile)

In this example, we do not even have to solve more than one equation at a time.