By Method of Joints
$\Sigma M_D = 0$
$6R_A = 4(1.2) + 3(2.4)$
$R_A = 2 ~ \text{kN}$
At Joint A
$\Sigma F_V = 0$
$\frac{4}{5}F_{AB} = 2$
$F_{AB} = 2.5 ~ \text{kN}$
$\Sigma F_H = 0$
$F_{AC} = \frac{3}{5}F_{AB}$
$F_{AC} = \frac{3}{5}(2.5)$
$F_{AC} = 1.5 ~ \text{kN}$
At Joint C
By inspection
$F_{BC} = 2.4 ~ \text{kN}$ and
$F_{CF} = 1.5 ~ \text{kN}$
At Joint F
$\Sigma F_H = 0$
$\frac{3}{5}F_{BF} = 1.5$
$F_{BF} = 2.5 ~ \text{kN compression}$ answer
Checking by Method of Sections
Joint E by inspection
$F_{BE} = 1.2 ~ \text{kN tension}$
For the section to the left of the cutting plane M-N
$\Sigma M_A = 0$
$6F_{BFv} = 4(1.2) + 3(2.4)$
$6F_{BF}(\frac{4}{5}) = 12$
$F_{BF} = 2.5 ~ \text{kN compression}$ answer
Comments
How did you take the
How did you take the direction of force on the joint ?
An easier to handle direction
An easier to handle direction of forces is by assuming all are tension (going away from joints). If your answer is positive then the assumption is correct. If the answer negative then the assumption is wrong, in means the correct direction is towards the joint (compression).