$\Sigma M_A = 0$

$3R_D = 50(2) + 80(0.75)$

$R_D = 53.33 \, \text{ kN}$

From the FBD of the section through a-a

$\Sigma M_E = 0$

$0.75F_{BC} + 2R_D = 0.75(80) + 1(50)$

$0.75F_{BC} + 2(53.33) = 60 + 50$

$F_{BC} = 4.45 \, \text{ kN tension}$ *answer*

$\Sigma M_C = 0$

$0.75F_{EF} = 1(R_D)$

$0.75F_{EF} = 53.33$

$F_{EF} = 71.11 \, \text{ kN tension}$ *answer*

$\Sigma F_V = 0$

$\frac{3}{5}F_{CE} + 50 = R_D$

$\frac{3}{5}F_{CE} + 50 = 53.33$

$F_{CE} = 5.55 \, \text{ kN tension}$ *answer*

## Comments

## How did it came up with 4&3

How did it came up with 4&3 instead of 0.75m and 1m?

## 0.75/1 = 3/4

0.75/1 = 3/4