Problem 424 - Method of Joints Checked by Method of Sections

$6R_A = 4(1.2) + 3(2.4)$

$R_A = 2 ~ \text{kN}$
 

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

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