Firstly, assume all unknowns are in tension, as shown in Figure
Next, make imaginary cuts around the joints, as shown by the circles in Figure (b). This action will give us three free body diagrams. The first we consider is around (1), because this has only two unknown forces; see Figure (c).
Resolving forces horizontally at ***** (1):
Forces to the left = forces to the right
i.e. F1 cos 30° = F2 cos 60°
i.e. 0.866 F1 = 0.5 F2
from which F1 =0.5 F2/0.866
i.e. F1 = 0.577 F2 ... (4.1)
Resolving forces vertically at ***** (1):
Upward forces = downward forces
i.e. 0 = 3 kN + F1 sin 30° + F2 sin 60°
i.e. 0 = 3 + 0.5 F1 + 0.866 F2 ....... (4.2)
Substituting equation (4.1) into equation (4.2) gives:
0 = 3 + 0.5 × 0.577 F2 + 0.866 F2
i.e. −3 = 1.1545 F2
from which, F2 = - 1/1.1545
i.e. F2 = −2.6 kN (compressive) ......(4.3)
Substituting equation (4.3) into equation (4.1) gives:
F1 = 0.577 × (−2.6)
i.e. F1 = −1.5 kN (compressive)
Consider next ***** (2), as it now has two or less unknown forces; see below Figure.
Resolving horizontally:
Forces to the left = forces to the right
i.e. 0 = F1 cos 30° + F3
However, F1 = −1.5 kN,
Hence, 0 = −1.5 × 0.866 + F3
from which, F3 = 1.30 kN(tensile)
These results are similar to those obtained by the graphical method and given in the table below.
| Member | Force (kN) |
| F1 | −1.5 |
| F2 | −2.6 |
| F3 | 1.3 |
